UC-NRLF 


^B    SE7    TS5 


PLANE   GEOMETRY 


WITH 


PROBLEMS  AND  APPLICATIONS 


BY 
H.    E.    SLAUGHT,   Ph.D. 

ASSOCIATE   PROFESSOR   OF   MATHEMATICS   IN    THE   UNIVERSITY 
OF  CHICAGO 

AND 

N.   J.    LENNES,    Ph.D. 

INSTRUCTOR   IN    MATHEMATICS   IN   THE   MASSACHUSETTS 
INSTITUTE   OF  TECHNOLOGY 


OF   THE 

UNIVERSITY 

OF 


Boston 

ALLYN    AND    BACON 

1910 


COPYRIGHT,   1910. 
BY   H.    E.    SLAUGHT 
AND   N.   J.   LENNES. 

7 


)Ci 


t/V- 


-7- 


Noriooofi  53reg8 
J.  8,  Cashing  Co.  —  Berwick  &  Smith  Co. 
•    Norwood,  Mass.,  U.S.A. 


PREFACE. 

In  writing  this  book  the  authors  have  been  guided  by  two 
main  purposes : 

(a)  That  pupils  may  gain  by  gradual  and  natural  processes 
the  power  and  the  habit  of  deductive  reasoning. 

(b)  That  pupils  may  learn  to  know  the  essential  facts  of 
elementary  geometry  as  properties  of  the  space  in  which  they 
live,  and  not  merely  as  statements  in  a  book. 

The  important  features  by  which  the  Plane  Geometry  seeks 
to  accomplish  these  purposes  are : 

1.  The  simpUJlcation  of  the  first  Jive  chapters  by  the  exclusion 
of  many  theorems  found  in  current  books.  These  five  chapters 
correspond  to  the  usual  five  books,  and  the  most  important 
omissions  are  the  formal  treatment  of  the  theory  of  limits, 
the  incommensurable  cases,  maxima  and  minima,  and  numer- 
ous other  theorems,  together  with  the  deduction  of  complicated 
algebraic  formulae,  such  as  the  area  of  a  triangle  and  the  radii 
of  the  inscribed,  escribed,  and  circumscribed  circles,  in  terms 
of  the  three  sides. 

Chapter  VI  contains  a  graphic  representation  of  certain  im- 
portant theorems  and  an  informal  presentation  of  incommen- 
surable cases  and  limits.  The  treatment  of  limits  is  based 
upon  the  graph,  since  the  visual  or  graphic  method  appeals 
more  directly  to  the  intuition  than  the  usual  abstract  processes. 
Chapter  VII  is  devoted  to  advanced  work  and  to  a  review  of 
the  preceding  chapters. 

iii 


IV  PREFACE. 

2.  Tfie  subject  has  been  enriched  by  including  many  applica- 
tions of  special  interest  to  pupils.  Here  an  effort  has  been  made 
to  include  only  such  concrete  problems  as  come  fairly  within 
the  observation  and  comprehension  of  the  average  pupil. 
This  led  to  the  omission,  for  example,  of  problems  relating 
to  machinery  and  technical  industries,  which  might  appeal 
to  an  exceptional  boy,  but  which  are  entirely  inappropriate 
for  the  average  student.  On  the  other  hand,  free  use  is  made 
of  certain  sources  of  problems  which  may  be  easily  compre- 
hended without  extended  explanation  and  which  involve  varied 
and  simple  combinations  of  geometric  forms.  Such  problems 
pertain  to  decoration,  ornamental  designs,  and  architectural 
forms.  They  are  found  in  tile  patterns,  parquet  floors,  lino- 
leums, wall  papers,  steel  ceilings,  grill  work,  ornamental 
windows,  etc.,  and  they  furnish  a  large  variety  of  simple 
exercises  both  for  geometric  construction  and  proofs  and  for 
algebraic  computation.  They  are  not  of  the  puzzle  type,  but 
require  a  thorough  acquaintance  with  geometric  facts  and 
develop  the  power  to  use  mathematics. 

These  problems  form  an  entirely  new  type  of  exercises,  and 
while  they  require  more  space  in  the  text-book  than  the  more 
difficult  "originals"  stated  in  the  usual  abstract  terms,  they 
excel  the  latter  in  interest  for  the  pupil  and  in  helping  to 
train  his  mathematical  common  sense.  Many  of  these  exer- 
cises are  simple  enough  to  be  solved  at  sight,  and  such  solution 
should  be  encouraged  whenever  possible.  All  the  designs  are 
taken  from  photographs  or  from  actual  commercial  patterns 
now  in  use.  By  thus  showing  that  the  abstract  theorems  of 
geometry  find  concrete  expression  in  a  multitude  of  familiar 
objects,  it  is  sought  to  make  the  subject  a  permanent  part 
of  the  pupil's  mental  equipment. 

3.  Persistent  effort  is  made  to  vitalize  the  content  of  the  defini- 
tions and  theorems.  It  is  well  known  that  pupils  often  study 
and  recite  definitions  and  theorems  without  really  comprehend- 


PREFACE.  V 

ing  their  meaning.  It  is  sought  to  check  this  tendency  by- 
giving  definitions  only  when  they  are  to  be  used,  and  by  imme- 
diately verifying  both  definitions  and  theorems  in  concrete 
cases.  The  figure  on  page  4  is  the  basis  for  a  large  number 
of  questions  of  this  type.  For  example,  see  §  25,  Ex.  3 ;  §  30, 
Ex.  1 ;   §  34,  Exs.  1,  2 ;  §  36,  Ex.  3 ;  §§  322,  324. 

In  this  connection  special  attention  is  called  to  the  emphasis 
placed  upon  those  theorems  which  are  of  fundamental  impor- 
tance both  in  the  logical  chain  and  in  their  immediate  use 
in  effecting  Constructions  and  indirect  measurements  otherwise 
difficult  or  impossible.  For  example,  see  the  theorems  on  con- 
gruence of  triangles,  §§  31-43,  the  constructions  of  §§  44-58, 
and  the  theorems  on  proportional  segments,  §§  243-254.  Com- 
pare especially  §  34,  Ex.  5,  §  244,  Ex.  2,  and  §  254,  Exs.  4,  5. 

The  summaries  at  the  close  of  the  chapters,  which  are  to  be 
made  by  the  pupil  himself,  will  vitalize  the  theorems  as  no 
made-to-order  summaries  can  possibly  do. 

4.  TJie  student  is  made  to  ajyproach  the  formal  logic  of  geometry 
by  natural  and  gradual  processes.  He  is  expected  to  gi^ow  into 
this  new,  and  to  him  unusual,  way  of  thinking.  The  treat- 
ment is  at  the  start  informal,  leading  through  the  congruence 
theorems  directly  to  concrete  applications  and  geometric  con- 
structions. The  formal  development  then  follows  gradually 
and  is  characterized  by  a  judicious  guidance  of  the  student, 
by  questions,  outlines,  and  other  devices,  into  an  attitude  of 
mental  independence  and  an  appreciation  of  clear  reasoning. 

This  informality  of  treatment,  most  frequent  in  the  earlier 
parts,  is  used  throughout  wherever  occasion  seems  to  justify 
it.  See  §§  318,  319.  An  effort  has  been  made  to  vary  the 
methods  of  attack  and  to  avoid  monotony.  Some  theo- 
rems are  proved  in  full;  some  are  outlined;  in  some,  hints 
and  suggestions  are  given.  Any  uniform  method  would  make 
it  impossible  to  leave  that  to  the  pupil  which  he  can  do  for 
himself,  and  at  the  same  time  to  give  full  assistance  where 


VI  PREFACE. 

that  is  needed.     In  no  case  are  questions  asked  whose  answers 
are  implied  by  the  form  of  the  questions. 

The  arrangement  of  the  text  is  adapted  to  three  grades  of 
courses : 

(a)  A  minimum  course,  consisting  of  Chapters  I  to  VI, 
without  the  problems  and  applications  at  the  end  of  each 
chapter.  This  would  provide  about  as  much  material,  theo- 
rems, constructions,  and  originals  as  is  found  in  the  briefest 
books  now  in  use. 

(6)  A  medium  course,  consisting  of  Chapters  I  to  VI,  includ- 
ing a  reasonable  number,  say  one  half  or  two  thirds,  of  the 
applications  at  the  end  of  each  chapter.  This  would  fully 
cover  the  college  entrance  requirements. 

(c)  An  extended  course,  including  Chapter  VII,  which  con- 
tains a  complete  review,  together  with  many  additional  theorems 
and  a  large  number  of  further  applications.  This  would 
provide  ample  work  for  the  strongest  high  schools,  and  for 
normal  schools  in  which  more  mature  students  are  found  or 
piore  time  can  be  given  to  the  subject. 

Chapter  VII  gives  a  complete  treatment  of  the  incommen- 
surable cases,  though  not  based  on  the  formal  theory  of  limits. 
It  is  believed  that  for  high  school  pupils  the  notion  of  a  limit  is 
best  studied  as  a  process  of  approximation,  and  that  the  best 
preparation  for  the  later  understanding  of  the  theory  is  by 
a  preliminary  study  of  what  is  meant  by  "  approaches,"  such 
as  is  given  in  Chapters  III  and  IV. 

Acknowledgment  is  due  to  Miss  Mabel  Sykes,  of  Chicago, 
for  the  use  of  a  large  number  of  drawings  and  designs  from 
her  extensive  collection ;  also  to  numerous  commercial  and 
manufacturing  houses,  both  in  this  country  and  in  Europe, 
through  whose  courtesy  many  of  the  patterns  were  obtained. 

H.    E.    SLAUGHT. 
N.   J.    LENNES. 
Chicago  and  Boston,  January,  1910. 


CONTENTS. 

Chapter  I.     Rectilinear  Figures. 

PAGE 

Introduction 1 

Notation  for  Points  and  Lines 5 

Angles  and  their  Notation 7 

Triangles  and  their  Notation       . 10 

Congruence  of  Geometric  Figures 12 

Tests  for  Congruence  of  Triangles .14 

Construction  of  Geometric  Figures     .         .         .         .        .         .         .20 

Theorems  and  Demonstrations 26 

Inequalities  of  Parts  of  Triangles 32 

Theorems  on  Parallel  Lines        .         .         ...         .         .         .35 

Applications  of  Theorems  on  Parallels       .         .         .         .        .         .  40 

Other  Theorems  on  Triangles     .         . 43 

Determinations  of  Loci 50 

Theorems  on  Quadrilaterals 54 

Polygons .  66 

Symmetry 68 

Methods  of  Attack 72 

Summary  of  Chapter  I .        .         .75 

Problems  and  Applications 76 

Chapter  II.     Straight  Lines  and  Circles. 

Theorems  on  the  Circle 86 

Measurement  of  Angles 88 

Problems  and  Applications 92 

Angles  formed  by  Tangents,  Chords,  and  Secants     ....  95 

Summary  of  Chapter  II 106 

Problems  and  Applications         .         . 107 

Chapter  III.     Measurement  of  Straight  Line-segments. 

Approximate  Measurement 112 

Ratios  of  Line-segments 114 

Theorems  on  Proportional  Segments 116 

vii 


Vlll  CONTENTS, 

PAGE 

Similar  Polygons 124 

Computations  by  Means  of  Sines  of  Angles 136 

Problems  of  Construction 140 

Summary  of  Chapter  III    .         .         . 145 

Problems  and  Applications 146 

Chapter  IV.     Areas  of  Polygons. 

Areas  of  Rectangles    .         .        .       ' 154 

Areas  of  Polygons      .         * 156 

Problems  of  Construction 166 

Summary  of  Chapter  IV 169 

Problems  and  Applications         .        .        .        ...        .        .        .170 

Chapter  V.     Regular  Polygons  and  Circles. 

Regular  Polygons 176 

Problems  and  Applications 183 

Measurement  of  the  Circle .188 

Summary  of  Chapter  V 194 

Problems  and  Applications 195 

Chapter  VI.     Variable  Geometric  Magnitudes. 

Graphic  Representation 207 

Dependence  of  Variables    .         .         . 215 

Limit  of  a  Variable    .       ^ *    .  217 

Chapter  VII.     Review  and  Further  Applications. 

On  Definitions  and  Proofs 220 

Loci  Considerations    ..........  226 

Problems  and  Applications 226 

Further  Data  concerning  Triangles    . 230 

Problems  and  Applications 233 

The  Incommensurable  Cases 238 

Problems  and  Applications 245 

Further  Applications  of  Proportion 250 

Further  Properties  of  Triangles 259 

Problems  and  Applications .  264 

Maxima  and  Minima  ...........  268 

Index .         .        .        .         .         .  277 


pla:^e  geometry. 

CHAPTER   I. 

RECTILINEAR   FIGURES. 

INTRODUCTION. 

1.  Elementary  geometry  is  a  science  which  deals  with  the 
space  in  which  we  live.  It  begins  with  the  consideration 
of  certain  elements  of  this  space  which  are  called  points, 
lines,  planes,  solids,  angles,  triangles,  etc. 

Some  of  these  terms,  such  as  point,  line,  plane,  are  here 
used  without  being  defined  in  a  strictly  logical  sense. 
Their  meaning  is  made  clear  by  description  and  by  con- 
crete illustrations  like  the  following. 

2.  Certain  portions  of  space  are  occupied  by  objects 
which  we  call  physical  solids,  as,  for  instance,  an  ordinary 
brick.      That  which  separates  a  solid  from 

the  surrounding  space  is  called  its  surface.     A 

This  may  be  rough  or  smooth.     If  a  surface 

is  smooth  and  flat,  we  call  it  a  plane  surface. 

A  pressed  .brick  has  six  plane  surfaces  called  faces.     Two 

adjoining  faces  meet  in  an  edge.     Three  edges  meet  in  a 

corner. 

The  brick  is  bounded  by  its  six  faces.  Each  face  is 
bounded  by  four  edges,  and  each  edge  is  bounded  by  two 
corners. 

1 


2  PLANE  GEOMETRY. 

3.  If  instead  of  the  brick  we  think  merely  of  its  form  and 
magnitude,  we  get  a  notion  of  a  geometrical  solid,  which 
has  the  three  dimensions,  length,  breadth,  and  thickness. 

The  faces  of  this  ideal  solid  are  called  planes.  These 
are  flat  and  have  length  and  breadth,  but  no  thickness. 

The  edges  of  this  solid  are  called  lines.  They  are 
straight  and  have  length,  but  neither  breadth  nor  thick- 
ness. The  corners  of  this  solid  are  called  points.  They 
have  position,  but  neither  length,  breadth,  nor  thickness  ; 
that  is,  they  have  no  magnitude. 

4.  It  is  possible  to  think  of  these  concepts  quite  inde- 
pendently of  any  physical  solid.  Thus  we  speak  of  the 
line  of  sight  from  one  point  to  another ;  and  we  say  that 
light  travels  in  a  straight  line. 

The  term  straight  line  is  doubtless  connected  with  the  idea  of  a 
stretched  string.     Of  all  the  lines  which  may  be  con- 
ceived as  passing  through  two  fixed  points  that  one         .^'^^       ^^ 
is  said  to  be  straight  between  these  points  which 
corresponds  most  nearly  to  a  stretched  string. 

Likewise  a  plane  may  be  thought  of  as  straight  or  stretched  in  every 
direction,  so  that  a  straight  line  passing  through  any  two  of  its  points 
lies  wholly  in  the  plane. 

5.  If  one  of  two  intersecting  straight  lines  turns  about 
their  common  point  as  a   pivot,  the    lines 
will    continue  to  have  only  one  point   in 
common  until  all  at  once  they  will  coincide 
throughout  their  whole  length.     Hence, 

Two  straight  lines  cannot  have  more  than 
one  point  iji  common  unless  they  coincide  and 
are  the  same  line;  that  is,  two  points  determine  a  straight  line. 

This  would  not  be  so  if  the  lines  had  width,  as  may 
be  seen  by  examining  the  figures. 


RECTILINEAR   FIGURES,  3 

6.  EXERCISES. 

^  1.  How  does  a  carpenter  use  a  straight-edge  to  determine  whether 
a  surface  is  a  plane  ?  Do  you  know  of  any  surface  to  which  this  test 
will  apply  in  one  direction  but  not  in  all  directions  ? 

2.  What  tool  does  a  carpenter  use  in  reducing  an  uneven  surface 
to  a  plane  surface?    Why  is  the  tool  so  named? 

3.  If  two  points  of  a  straight  line  lie  in  a  plane,  what  can  be  said 
of  the  whole  line  ? 

4.  How  many  points  of  a  straight  line  can  lie  in  a  plane  if  it  con- 
tains at  least  one  point  not  in  the  plane  ? 

-  5.  If  two  straight  lines  coincide  in  more  than  one  point,  what  can 
you  say  of  them  throughout  their  whole  length  ?  Do  you  know  of 
any  lines  other  than  straight  lines  of  which  this  must  be  true  ? 

—  6.  How  do  the  material  points  and  lines  made  by  crayon  or  pencil 
differ  in  magnitude  from  the  ideal  points  and  lines  of  geometry  ? 

7.  A  machine  has  been  made  which  rules  20,000  distinct  lines  side 
by  side  within  the  space  of  one  inch.  Do  such  lines  have  width  ?  Are 
they  geometrical  lines  ? 

8.  Of  all  the  lines,  straight  or  curved,  through  two  points,  on 
which  one  is  the  shortest  distance  measured  between  the  two  points  ? 
See  the  figure  of  §  4. 

Historical  Note.  The  Egyptians  appear  to  have  been  the  first 
people  to  accumulate  any  considerable  body  of  exact  geometrical 
facts.  The  building  of  the  great  pyramids  (before  3000  B.C.)  re- 
quired not  a  little  knowledge  of  geometric  relations.  They  also 
used  geometry  in  surveying  land.  Thus  it  is  known  that  Rameses  II 
(about  1400  B.C.)  appointed  surveyors  to  measure  the  amount  of  land 
washed  away  by  the  Nile,  so  that  the  taxes  might  be  equalized. 

The  Greeks,  however,  were  the  first  to  study  geometry  from  a  logi- 
cal point  of  view.  Between  600  b.c,  M^hen  Thales,  a  Greek  from  Asia 
Minor,  learned  geometry  from  the  Egyptians,  and  300  B.C.,  when 
Euclid,  a  Greek  residing  in  Alexandria,  Egypt,  wrote  his  Elements  of 
Geometry,  the  crude,  practical  geometric  information  of  the  Egyp- 
tians was  transformed  into  a  well-nigh  perfect  logical  system. 

Euclid's  "  Elements  "  contains  the  essential  facts  of  every  text- 
book on  elementary  geometry  that  has  been  written  since  his  time. 


PLANE  GEOMETRY 


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RECTILINEAR   FIGURES.  6 

NOTATION  FOR  POINTS  AND  LINES. 

7.  A  point  is  denoted  by  a  capital  letter. 

A  straight  line  is  denoted  by  two  capital  ' 

letters  marking  two  of  its  points  or  by     — '^^^     ^,-? — 
one   small   letter.      The   word   line   alone  i 

usually  means  straight  line. 

Thus,  the  point  A,  the  line  AB,  or  the  line  /. 

8.  A   straight   line   is   usually   understood  to  be  un- 
limited in  length  in  both  directions,  while  that  part  of  it 
which  lies  between  twO  of   its   points   is  ^ 
called  a  line- segment,  or  simply  a  segment. 

These  points  are  called  the  end-points  of  the  segment. 
Thus,  the  segment  AB  or  the  segment  a. 
Two  segments  with  the  same  end-points  are  coincident. 

—.    9.   A  part  of  a  straight  line,  called  a  ray  or  half -line,  may 
be  thought  of   as  generated  by  a  point 
starting  from  a  fixed  position  and   mov-     '  '     ^ 

ing  indefinitely  in  one  direction.     The  starting  point  is 
called  the  end-point  or  origin  of  the  ray. 

If  A  is  the  origin  of  a  ray  and  B  any  other  point  on  it,  then  it  is 
read  the  ray  AB,  not  tJie  ray  BA. 

10.   Two  line-segments   are   said   to   be 

added  if  they  are  placed  end  to  end  so  as    4 £ ^ 

to  form  a  single  segment. 

Thus,  segment  ^C  =  segment  ^5  + segment  BC, or  AC —  AB+BC. 

If  AC  —  AB  -\-  BC,  then  AC  is  greater  than  either  AB  or 
BC,  and  this  is  written  AC  >  AB  and  AC  >  BC. 

A  segment  may  also  be  subtracted  from  a  greater  or 
from  an  equal  segment. 

Thus,  \iAC  =  AB-\-  BC,  then  AB=AC-  5Cand  BC  =  AC  -  AB. 


PLANE  GEOMETRY. 


Nl^      A  segment  is  multiplied  by  an  integer  n  by  taking  the 
sum  of  n  sucb  segments. 

Thus,  if  A  C  is  the  sum  of  n  segments  each  AB,  then  A  C  =  n  •  AB. 

If  AG  is  n  times  AB^  then  AC  may  be  divided  by  w. 

Thus,  AB  =  AC  -^  n  or  AB  =  -.  AC. 

n 

11.  A  broken  line  is  composed  of 
connected  line-segments  not  all  lying 
in  the  same  straight  line. 

A  curved  line,  or  simply  a  curve,  is 
a  line  no  part  of  which  is  straight. 

A  curved  line  or  a  broken  line  may 
inclose  a  portion  of  a  plane,  while  a 
straight  line  cannot. 

12.  A  circle  is  a  plane  curve  containing  all  points  equally 
distant  from  a  fixed  point  in  tlie  plane, 

/^   and  no  other  points. 

The  fixed  point  is  called  the  center  of 
the  circle.  Any  line-segment  joining 
the  center  to  a  point  on  the  circle  is  a 
radius  of  the  circle. 

Any  portion  of  a  circle  lying  between 
two  of  its  points  is  called  an  arc. 

Evidently  all  radii  of  the  same  circle  are  equal. 

Any  combination  of  points,  segments,  lines,  or  curves  in 
a  plane  is  called  a  plane  geometric  figure. 

Plane  Geometry  deals  with  plane  geometric  figures. 


13. 


EXERCISES. 


1.  How  many  end-points  has  a  straight  line?     How  many  has  a 
line-segment?     How  many  has  a  ray?     A  circle? 

2.  Can  you  inclose  a  portion  of  a  plane  with  two  line-segments? 
With  three?     With  four? 


BECTILINEAR  FIGUBES. 


ANGLES  AND  THEIR  NOTATION. 


14.  An  angle  is  a  figure  formed  by  two  rays  proceeding 
from  the  same  point.     The  point  is  the 
vertex  of  the  angle  and  the  rays  are  its 
sides.     The  angle  formed  by  two  rays  is   ^^^^"^^       sHT 
said  to  be  the  angle  between  them  or  simply  their  angle. 

Two  line-segments  having  a  common  end-point  also  form  an  angle, 
namely,  the  angle  of  the  rays  on  which  the  segments  lie.  An  angle  is 
determined  entirely  by  the  relative  directions  of  its  rays  and  not  by 
the  lengths  of  the  segments  laid  off  on  them. 

15.  An  angle  is  denoted  by  three  letters,  one  at  its 
vertex  and  one  marking  a  point  on  each  of  its  sides.  The 
one  at  the  vertex  is.  read  between  the  other  ^ 
two,  as  the  angle  CAB^  or  the  angle  BAC^ 
not  ABC.  The  one  letter  at  the  vertex  is 
also  used  alone  to  denote  an  angle  in  case  ^ 
no  other  angle  has  the  same  vertex,  as,  for  instance,  the 
angle  A. 

In  case  several  angles   have  the  same  vertex,  a  small 
letter  or  figure  placed  within  each  angle, 
together  with    an   arc    connecting   its 
sides,  is  a  convenient  notation.      The 
sign  Z  is  used  for  the  word  angle. 

Thus  in  the  figure  we  have  Zl,  Z2,  Z  3,  read, 
angle  one,  angle  two,  angle  three.  a 

16.  If  two  rays  lie  in  the  same   straight  line  and  ex- 
tend  from    the  same  end-point   in  opposite  s^ 
directions  they  are  said  to  form  a  straight  | 
angle.      If   they  extend  in  the  same  direc-  side    >  side 
tion,  they  coincide  and  form  a  zero  angle. 


8  PLANE  GEOMETRY. 

17.  An  angle  may  be  thought  of  as  generated  by  a  ray 
turning  about  its  end-point  as  a  pivot. 

Thus  Z  BA  C  is  generated  by  a  ray  rotating 
from  the  position  AB  to  the  position  A  C.  The 
rotating  ray  is  usually  conceived  as  moving 
in  the  direction  opposite  to  the  hands  of  a 
clock  and  the  sides  of  the  angle  should  usu- 
ally be  read  in  this  order.  Thus  ZBAC,  not 
ZCAB. 

If  the  ray  continues  to  rotate  until  ^-^ 

it  lies  in  a  direction  exactly  opposite         — — '^  *    ' 
to  its  original  position,  it  generates  a 
straight  angle,  as   the  straight  angle 
BAC. 

If  the  ray  rotates  until  it  reaches 
its  original  position,  the  angle  generated  is  called  a  peri- 
gon,  that  is,  an  angle  of  complete  rotation.  ' 

The  position  from  which  the  rotating  ray  starts  is  called  the  origin 
of  the  angle.  From  this  point  of  view  two  rays  from  the  same  point 
form  two  angles  according  as  one  or  the  other  of  the  rays  is  regarded  as 
the  origin.  In  elementary  geometry  only  angles  less  than  or  equal  to 
a  straight  angle  are  usually  considered. 

The  units  of  measure  for  angles  are  one  three-hun- 
dred-sixtieth of  a  perigon  which  is  called  a  degree,  one 
sixtieth  of  a  degree  called  a  minute,  and  one  sixtieth  of  a 
minute  called  a  second.  These  are  denoted  respectively  b}^ 
the  symbols  °,  ',  '^  Thus,  an  angle  of  20°  45'  30".  A 
straight  angle  is  therefore  an  angle  of  180°  and  a  perigon 
is  an  angle  of  360°. 

18.  Angles  are  measured  by  means  of  an  instrument 
called  a  protractor,  which  consists  of  a  semicircular  scale 
with  degrees  from  0°  to  180°  marked  upon  it. 

An  inexpensive  protractor  made  of  cardboard  or  brass  may  be 
had  at  any  stationery  store.     See  the  figure  of  §  33. 


RECTILINEAR  FIGURES.  9 

19.  Two  angles  are  said  to  be  equal  if  they  can  be  made 
to  coincide  without  changing  the  form  of  either. 

If  a  ray  is  drawn  from  a  point  in  a  §a 

straight  line  so  that  the  two  angles  thus 
formed   are   equal,  each  angle  is  called  /^ 

a  right  angle,  and  the  ray  is  said  to  be   ^ 

perpendicular  to  the  line. 

Thus,  ii  Zl  =  Z2,  each  angle  is  a  right  angle  and  AC  is  then  said 
to  be  perpendicular  to  BD. 

Since  the  straight  angle  BAB  is  composed  of  Z  1  and 
Z  2,  each  of  which  is  a  right  angle,  it  appears  that 
a  straight  angle  equals  two  right  angles. 

See  §  39  for  the  addition  of  angles  in  general. 

An  acute  angle  is  less  than  a  right  angle. 

An  obtuse  angle  is  greater  than  a  right 


./ 


angle  and  less  than  a  straight  angle.  ^Xc^^^^ 


N.v^^'' 


Acute  and  obtuse  angles  are  called  oblique 
angles. 

A  reflex  angle  is  greater  than  a  straight 
angle  and  less  than  a  perigon. 

One  line  is  oblique  to  another  if  the  angles 
between  them  are  oblique. 

A  ray  which  divides  an  angle  into  two    ^        2" 
equal  angles  is  called  its  bisector.     Thus  a  perpendicular 
is  the  bisector  of  a  straight  angle. 

20.  EXERCISES. 

1.    Since  we  can  always  place  two  straight  angles  so  as  to  make 
them  coincide,  what  can  we  say  as  to  whether  or  not  they  are  equal? 
What  of  two  right  angles? 
^^^  2.   What  part  of  a  straight  angle  is  a  right  angle? 

3,    How  many  degrees  in  a  right  angle?     In  two  right   angles? 
What  kind  of  an  angle  is  one  of  45°?     One  of  130°? 


10  PLANE  GEOMETRY. 

4.  Suppose  that  in  the  figure  the  ray  AC  rotates  about  the 
point  A  from  the  position  AB  to  the  position  AD.  What  change 
takes  place  in  Zl?  What  in  Z2?  Can  there  be 
more  than  one  position  oi  AC  for  which  Z  1  =  Z 2 ? 
In  this  way  it  may  be  made  clear  that  any  angle 
has  one  and  only  one  bisector.  jT         ^  b 

5.  How  many  rays  perpendicular  to  BD  at  the  point  A  can  be 
drawn  on  the  same  side  of  BDl  Does  the  answer  to  this  question 
depend  upon  the  answers  to  the  questions  in  Ex.  4  ?     How  ? 

6.  Pick  out  three  acute  angles,  three  right  angles,  and  three  obtuse 
angles  in  the  figure  on  page  4. 

TRIANGLES   AND   THEIR   NOTATION. 

21,  If  the  points  A^  1?,  C  do  not  lie  in  the  same  straight 
line,  the  figure  formed  by  the  three  segments,  AB^  BC,  and 
CA^  is  called  a  triangle. 

The  segments  are  the  sides  of  the  triangle,  and  the 
points  are  its  vertices.     The  symbol  A  is  b 

used  for  the  word  triangle. 

Each  angle  of  a  triangle  has  one  side 
opposite    and    two   sides    adjacent   to   it.  &       . 

Similarly  each  side  of  a  triangle  has  one  angle  opposite 
and  two  angles  adjacent  to  it.  .  The  side  opposite  an  angle 
is  often  denoted  by  the  corresponding  small  letter. 

Thus,  in  the  figure  the  side  a  and  the  angle  A  are  opposite  parts,  as 
are  angle  B  and  side  b  and  angle  C  and  side  c.  The  three  sides  and 
three  angles  of  a  triangle  are  called  the  parts  of  the  triangle. 

These  six  parts  are  considered  as  lying  in  order  around  the  figure, 
an  Z  A,  side  b,  Z  C,  side  a,  etc. 

An  angle  of  a  triangle  is  said  to  be  included  between  its 
two  adjacent  sides,  and  a  side  is  said  to  be  included  be- 
tween the  two  angles  adjacent  to  it. 

Thus,  in  the  figure  the  side  a  is  included  between  Z  B  and  Z  C,  and 
ZA  is  included  between  the  sides  b  and  c. 


BECTILINEAR   FIGURES. 


11 


22.   A  triangle  is  called  equilateral  if  it  has  its  three 
sides  equal,  isosceles  if  it  has  at  least  two  sides  equal, 


scalene  if  it  has  no  two  sides  equal,  equiangular  if  it  has  its 
three  angles  equal. 

Select  each  kind  from  the  figures  on  this  page. 

23.  A  triangle  is  called  a  right  triangle  if   it  has  one 
right  angle,  an  obtuse  triangle  if  it  has  one 
obtuse   angle,   an    acute    triangle   if    all   its 
angles  are  acute. 

Select  each  kind  from  the  figures  on  this  page. 

The  side  of  a  right  triangle  opposite  the 
right  angle  is  called  the  hypotenuse  in  dis- 
tinction from  the  other  two  sides,  which  are  sometimes 
called  its  legs. 

24.  The  side  of  a  triangle  on  which  it  is  supposed  to 
stand  is  called  its  base.  The  angle  opposite  the  base  is 
called  the  vertex  angle, 
and  its  vertex  is  the 
vertex  of  the  triangle. 

The  altitude  of  a  tri- 
angle is  the  perpendic- 
ular from  the  vertex  to 
the  base  or  the  base  produced.  Evidently  any  side  may 
be  taken  as  the  base,  and  hence  a  triangle  has  three  different 
altitudes. 


Vertex 


Vertex 


Base 


12  PLANE  GEOMETRY. 

25.  EXERCISES. 

1.  Is  every  equilateral  triangle  also  isosceles?  Is  every  isosceles 
triangle  also  equilateral  ? 

i 

2.  Is  a  right  triangle  ever  isosceles?    Is  an  obtuse  triangle  ever 

isosceles?    Draw  figures  to  illustrate  your  answers. 

3.  In  the  figure  on  page  4  determine  by  measuring  sides  which  of 
the  triangles  HNP,  LKW,  IHN,  MIJ,  KVU,  OKJ,  LVW,  are  isos- 
celes, which  are  equilateral,  and  which  are  scalene. 

4.  Determine  whether  /,  K,  V  of  the  same  figure  may  be  the 
vertices  of  a  triangle ;  also  whether  J,  0,  G  may  be. 

5.  Pick  out  ten  obtuse  triangles  in  this  figure;  also  ten  acute 
triangles. 

CONGRUENCE  OF  GEOMETRIC  FIGURES. 

26.  In  comparing  geometric  figures  it  is  assumed  that 
they  may  he  moved  about  at  will^  either  in  the  same  plane  or 
out  of  it,  without  changing  their  shape  or  size. 

27.  Two  figures  are  said  to  be  similar  if  they  have  the 
same  shape.  This  is  denoted  by  the  symbol  ~,  read  is 
similar  to. 

For  a  more  precise  definition  see  §§  255,  256. 

Two  figures  are  said  to  be  equivalent  or  simply  equal  if 
they   have    the    same    size   or    magnitude. 
This  is  denoted  by  the  symbol  =,  read  is    i — i 
equivalent  to  or  is  equal  to.  L I 

Two  figures  are  said  to  be  congruent  if,    | — . 

without  changing  the  shape  or  size  of  either,     I I  ~ 

they   may   be    so    placed    as    to    coincide    . ,, . 

throughout.     This  is  denoted  by  the  symbol  ~ 

^,  read  is  congruent  to. 

In  the  case  of  line-segments  and  angles,  congruence  is  determined 
by  size  alone.  Hence  hi  these  cases  we  use  the  symbol  —  to  denote 
congruence,  and  read  it  equals  or  is  equal  to. 


RECTILINEAR  FIGURES. 


13 


28^  It  is  clear  that  if  each  of  two  figures  is  congruent  to 
the  -^ame  figure  they  are  congruent  to  each  other. 

Hence  if  we  make  a  pattern  of  a  figure,  say  on  tracing 
paper,  and  then  make  a  second  figure  from  this  pattern, 
the  two  figures  are  congruent  to  each  other. 


29.    li  AABC  ^AA^b'c' ,  the  notation  of  the  triangles 

c 


may    be     so     arranged     that     AB  =  A'b\ 


bc  =  b'c',  ca  =  c'a',  z_^— ^^  , 
and  ZC=Zc'.  In  this  case  AB  is  said  to 
correspond  to  a'b',  BC  to  b'c',  CA  to  c'a\ 
ZA  to  Za',  etc. 

Hence,  we  say  that  corresponding  parts  of 
congruent  triangles  are  equal. 


30. 


EXERCISES. 


1.  Using  tracing  paper,  draw  triangles  congruent  to  the  triangles 
MIN,  NHP,  OAB,  OFE,  OKL,  UKV,  OGL  on  page  4,  and  by  ap- 
plying the  pattern  of  each  triangle  to  each  of  the  others  determine 
whether  any  two  are  congruent. 


2.  Find  as  in  §  28  whether  any  two  of  three  accompanying  triangles 
are  congruent,  and  if  so  arrange  the  notation  so  as  to  show  the  corre- 
sponding parts. 

3.  Give  examples  of  figures  which  are  similar,  equal,  or  con- 
gruent, different  from  those  in  §  27. 

_.   4.   If  two  figures  are  congruent,  does  it  follow  that  they  are  equal? 
Similar  ? 

5.  If  two  figures  are  similar,  does  it  follow  that  they  are  equal? 
Congruent? 

.^ 6.   If  two  figures  are  equal,  are  they  similar?     Congruent? 


14  PLANE  GEOMETBY, 

TESTS  FOR  CONGRUENCE  OF  TRIANGLES. 

31.  The  method  of  determining  whether  two  triangles 
are  congruent  by  making  a  pattern  of  one  and  applying  it 
to  the  other  is  often  inconvenient  or  impossible.  There 
are  other  methods  in  which  it  is  necessary  only  to  determine 
whether  certain  sides  and  angles  are  equal. 

These  methods  are  based  upon  three  important  tests  for 
congruence  of  triangles. 

32.  First  Test  for  Congruence  of  Triangles. 

If  two  triangles  have  two  sides  and  the  included  angle 
of  one  equal  respectively  to  tivo  sides  and  the  included 
'  a7igle  of  the  other,  the  triangles  are  congruent. 
■  This  may  be  shown  by  the  following  argument: 


Let  ABC  and  A'BC  be  two  triangles  in  which  AB  =  A^B!, 
AC  =  A'C,  and  Z.A  =  ZA'. 

We  are  to  show  that  A ^5C^  A ^'s'c'. 

Place  A  ABC  upon  AA'b'c'  so  that  Z.A  coincides  with 
Z^',  which  can  be  done  since  it  is  given  that  Z.a=Za', 

Then  point  B  will  coiYicide  with  b'  and  G  with  c',  since 
it  is  given  that  AB  =  A'b'  and  AC  =  A'c', 

Hence,  side  BC  will  coincide  with  b'c'  (§  8). 

Thus,  the  two  triangles  coincide  throughout  and  hence 
are  congruent  (§  27).  '  ^' 


The  process  just  used  is  called  superposition.    It  ^ 
may  sometimes  be  necessary  to  move  a  figure  out 
of  its  plane  in  order  to  superpose  it  upon  another 
as  in  the  case  of  the  accompanying  triangles. 


BECTILINEAB  FIGURES.  15 

33.  The  equality  of  short  line-segments  is  conveniently 
tested  by  means  of  the  dividers  or  compasses. 

Place  the  divider  points  on  the  end-points  of  one  segment  ^5  and 
then  see  whether  they  will  also  coincide  with  the  end-points  of  the 
other  segment  A'B'.     If  so,  the  two  segments  are  equal. 

The  equality  of  two  angles  may  be  tested  by  means  of 
the  protractor. 

Place  the  protractor  on  one  angle  i^OC  as  shown  in  the  figure  and 
read  the  scale  where  OC  crosses  it.     Then 
place    the   protractor    on   the    other  angle       y^^^^*'""'"'"'"^*^^^ 
B'O'C  and  see  whether  O'C  crosses  the  scale 
at  the  same  point.    If  so,  the  two  angles  are 
equal. 

34.  EXERCISES. 

1.  Using  the  protractor  determine  which  pairs  of  the  following 
angles  on  page  4  are  equal : 

HPG,  LGW,  GWL,  AOB,  VLW,  LVW. 

2.  By  the  test  of  §  32  determine  whether,  on  page  4, 

AJKU^AG  WL,  also  whetiier  A  MIH  ^AKVW. 
First  find  whether  two  sides  of  one  are  equal  respectively  to  two 
sides  of  the  other,  and  if  so  compare  the  included  angles. 

3.  Could  two  sides  of  one  triangle  be  equal  respectively  to  two 
sides  of  another  and  still  the  triangles  not  be  congruent  ?  Illustrate 
by  constructing  two  such  triangles. 

4.  Show  by  the  test  of  §  32  that  two  right  triangles  are  congruent 
if  the  legs  of  one  are  equal  respectively  to  the  legs  of  the  other.  Can 
this  be  shown  directly  by  superposition  ? 

-     5.   Find  the  distance  AB  when,  on   account    of 
some  obstruction,  it  cannot  be  measured  directly. 

Solution.  To  some  convenient  point  C  measure 
the  distances  A  C  and  BC.  Continuing  in  the  direc- 
tion A C  lay  off  CA'  =  AC,  and  in  the  direction  BC 
lay  off  CB'  =  BC.  Then  Z 1  =  Z 2  (see  §  74) .  Test 
this  with  the  protractor.  Show  that  the  length  AB 
is  found  by  measuring  A'B'. 


16 


PLANE  GEOMETRY. 


35.  Second  Test  for  Congruence  of  Triangles. 

If  tiDO  triangles  have  two  angles  and  the  included 
side  of  one  equal  respectively  to  two  angles  and  the  in- 
cluded side  of  the  other ^  the  triangles  are  congruent 

This  is  shown  by  the  following  argument ; 

G  \cj  ' 


A  B  A'  B' 

Let  ABC  and  A^B'C  be  tw^  triangles  in  which  Z.A=:i/-A\ 
Z.B=^ZEf,  and  AB=A'Ef.  <« 

We  are  to  show  that  AABC^Aa'b'c'  . 

Place  A  ABC  upon  AA'b'c'  so  that  AB  coincides  with 
its  equal  A'b',  making  C  fall  on  the  same  side  oi  a' B^  as  c'. 

Then  ^C  will  take  the  direction  of  A'&,  since  ZA  =  Z j.', 
and  the  point  C  must  fall  somewhere  on  the  ray  a'&. 

Also  BC  will  take  the  direction  of  b'c'  (Why?),  and 
hence  C  must  lie  on  the  ray  b'c'. 

Since  the  point  C  lies  on  both  of  the  rays  A'&  and  b'c'^ 
it  must  lie  at  their  point  of  intersection  &  (§5).  Hence, 
the  triangles  coincide  and  are,  therefore,  congruent  (§  27). 

36.  EXERCISES. 

1.  In  the  figure  of  §  35  is  it  necessary  to  move 
A  ABC  out  of  the  plane  in  which  the  triangles 
lie  ?    Is  it  necessary  in  the  figure  here  given  ? 

2.  Show  how"  to  measure  the  height  of  a  tree 
by  using  the  second  test  for  congruence. 

Suggestion.  Lay  out  a  triangle  on  the  ground 
which  is  congruent  to  A  ABC,  using  §  35. 

3.  By  the  second  test  determine  whether 
A  OHG  ^  A  OJK  on  page  4. 

4.  Draw  any  triangle.     Construct  another  tri- 


RECTILINEAR  FIGURES. 


17 


angle  congruent  to  it.  Use  §  35  and  also  §  32. 
Use  the  protractor  to  construct  the  angles. 

5.  Find  the  distance  A  C,  when  C  is  inaccessible. 
Let  i^  be  a  convenient  point  from  which  A  and 

C  are  visible.  Lay  ont  a  triangle  ABC  mak- 
ing Z  3  =  Z  1  and  Z  4  =  Z  2.  Show  that  the  dis- 
tance AC  may  be  found  by  measuring  AC. 

6.  Show  how  to  find  the  distance  between  two 
inaccessible  points  A  and  B. 

Solution.  Suppose  that  both  A  and  B  are 
visible  from  C  and  D.  (1)  Using  the 
triangle  CDA,  find  the  length  of  AD  as  in 
Ex.  5  above.  (2)  Using  the  triangle  CBD, 
find  DB  in  the  same  manner.  (3)  Using 
the  triangle  DBA,  find  ^i?  as  in  Ex.  5,  §  34. 

37.  The  proof  of  the  third  test  for 
congruence  of  triangles  involves  the 
following : 

The  angles  opposite  the  equal  sides  of  an  isosceles  tri 
angle  are  equal. 

Let  ABC  be  an  isosceles  triangle  having 
AC  =  EC 

We  are  to  show  that  Z  A  =  Z  b. 

Suppose  CD  divides  z.  acb  so  that 
Z1=Z2. 

By  means  of  §  32  show  that 
A  ACD  ^AbCD. 

Then   Z  ^  =  Z  7;  by  §  29. 

The  theorems  §  35  and  §  37  are  due  to  Thales.     It  is  said  he  used 
§  35  in  calculating  the  distance  from  the  shore /^o  a  ship  at  sea. ' 


38. 


EXERCISE. 


On  page  4  pick  out  as  many  pairs  of  angles  as  possible  which  may 
be  shown  to  be  equal  by  §  37.     Test  these  by  using  the  protractor. 


18 


PLANE  GEOMETRY. 


39.  Definitions.  Two  angles  which  have  a 
common  vertex  and  a  common  side  are  said 
to  be  adjacent  if  neither  angle  lies  within 
the  other.  /^  i\ 

Thus,  Z  1  and  Z.  2  are  adjacent,  while  Z 1  and  Z  3  are  not  adjacent. 

The  sum  of  two  angles  is  the  angle  formed  by  the  sides 
not  common  when  the  two  angles  are  placed  adjacent. 

Thus,  Z3  =  Z1  +  Z2. 

IfZ3  =  Zl  +  Zl2,  then  we  say  that  Z 3  is  greater  than 
either  Z 1  or  Z  2.     This  is  written  Z  3  >  Z 1  and  Z  3  >  Z  2. 

An  angle  may  also  be  subtracted  from  a  greater  or  equal 
angle.  ThusifZ3  =  Zl  +  Z  2,  then  Z3  -Zl  =  Z2and 
Z3  —  Z  2  =  Z 1.     It  is  clear  that  : 

If  equal  angles  are  added  to  equal  angles^  the  sums  are 
equal  angles. 

Angles  may  be  multiplied  or  divided  by  a  positive  in- 
teger as  in  the  case  of  line-segments.     See  §  10. 

40.  We  may  now  prove  the  third  test  for  congruence  of 
triangles,  namely : 

//  two  triangles  have  three  sides  of  one  equal  respec- 
tively to  three  sides  of  the  other,  the  triangles  are  con- 
gruent* ' 


Let    ABC  and  A'B'C' he  two  triangles  in  which  AB  =  A'B'j 
BC=B'C',CA  =  CA'. 

We  are  to  show  that  A  ABC  ^  AA'b'c', 


RECTILINEAR  FIGURES.  19 

Place  Aa'b'c'  so  that  A'b'  coincides  with  AB  and  so 
that  c'  falls  on  the  side  of  AB  which  is  opposite  C. 
(Why  is  it  possible  to  make  A'B'  coincide  with  AB"}) 

Draw  the  segment  CC'.  From  the  data  given,  how  can 
§  37  be  used  to  show  that  in  A  ACC'  Z.1  =  Z  2  ? 

Use  the  same  argument  to  show  that  Z  3  =  Z  4. 

But  if  Z  1  =  Z  2 

and  Z3  =  Z4, 

then  Z  1  +  Z  3  =  Z  2  +  Z  4.  (§39) 

That  is,  ^ACB  =  Z  BC'A. 

How  does  it  now  follow  that  A  ABC  ^  A  ABC'  ?    (§  32) 

But  AABC'  ^AA'b'c'.  (§26) 

Hence,  AABC^Aa'b'c'.  (§28) 

Make  an  outline  of  the  steps  in  the  above  argument,  and  see  that 
each  step  is  needed  in  deriving  the  next. 

41.  Definition.  If  one  triangle  is  congruent  to  another 
because  certain  parts  of  one  are  equal  to  the  corresponding 
parts  of  the  other,  then  these  parts  are  said  to  determine 
the  triangle.  That  is,  ant/  other  triangle  constructed  with 
these  given  parts  will  be  congruent  to  the  given  triangle. 


42. 


EXERCISES. 


1.  In  §  37  show  that  CD  is  perpendicular  to  AB  and  that  AD  =  DB. 
State  this  fully  in  words. 

2.  Using  §  40,  determine  which  of  the  following  triangles  on 
page  4  are  congruent :  OJK,  HNP,  OTH,  PRG,  JKU. 

3.  Do  two  sides  ^e^ermme  a  triangle ?  Three  sides?  Two  angles? 
Three  angles?    Illustrate  by  figures. 

4.  A  segment  drawn  from  the  vertex  of  an  isosceles  triangle  to  the 
middle  point  of  the  base  bisects  the  vertex  angle  and  is  perpendicular 
to  the  base. 

5.  What  parts  of  a  triangle  have  been  found  sufficient  to  determine 
it?    In  each  case  how  many  parts  are  needed? 


20  PLANE  GEOMETRY. 

43.  The  three  tests  for  congruence  of  triangles,  §§  32, 
35,  40,  lie  at  the  foundation  of  the  mathematics  used  in 
land  surveying.  The  fact  that  certain  parts  of  a  triangle 
determine  it  shows  that  it  mai/  be  possible  to  compute  the 
other  parts  when  these  parts  are  known.  Rules  for  doing 
this  are  found  in  Chapter  III. 

CONSTRUCTION   OF   GEOMETRIC   FIGURES. 

44.  The  straight-edge  ruler  and  the  compasses  are  the 
instruments  most  commonly  used  in  the  construction  of 
geometric  figures. 

By  means  of  th^  ruler  straight  lines  are  drawn,  and  the 
compasses  are  used  in  laying  off  equal  line-segments  and 
also  in  constructing  arcs  of  circles  (§  12). 

Other  common  instruments  are  the  protractor  (§  33)  and 
the  triangular  ruler  with  one  square  corner  or  right  angle. 

The  three  tests  for  congruence  of  two  triangles  are  of 
constant  use  in  geometrical  constructions. 

45.  Peoblem.    To  find  a  point  ivhose  distances  from 
the  extremities  of  a  given  seg-  \g/^ 
ment  are  specified. 

Solution.    Let  AB  be  the  given  seg- 
ment and  let  it  be  required  to  find  a   A b 

point  C  which  shall  be  one  inch  from  each  extremity  of  AB. 

Set  the  points  of  the  compasses  one  inch  apart.  With 
^  as  a  center  draw  an  arc  m^  and  with  S  as  a  center  draw 
an  arc  n  meeting  the  arc  m  in  the  point  C.  Then  every 
point  in  the  arc  m  is  one  inch  from  A  and  every  point  in 
the  arc  n  is  one  inch  from  B  (§  12). 

Hence  C,  which  lies  on  both  m  and  w,  is  one  inch  from  A 
and  also  from  B. 


RECTILINEAR   FIGURES.  21 

46.  EXERCISES. 

1.  In  the  preceding  problem  is  there  any  other  point  in  the  plane 
besides  C  which  is  one  inch  distant  from  both  A  and  B'i  If  so, 
show  how  to  find  it. 

2.  Could  AB  be  given  of  such  length  as  to  make  the  construction 
in  §  45  impossible  ? 

3.  Is  there  any  condition  under  which  one  point  only  could  be 
found  in  the  above  construction  ?  If  so,  what  would  be  the  length  of 
AB'} 

4.  Find  a  point  one  inch  from  A  and  two  inches  from  B  and  dis- 
cuss all  possibilities  as  above. 

5.  Given  three  segments  a,  6,  c,  construct  a  triangle  having  its  sides 
equal  to  these  segments.  Discuss  all  possibilities  depending  upon 
the  relative  lengths  of  the  given  segments. 

47.  Problem.  To  construct  an  angle  equal  to  a 
given  angle,  ivithout  using  the  py^otractor. 


Solution.    Given  the  angle  A. 

Lay  off  any  distance  AB  on  one  of  its  sides  and  any  dis- 
tance AC  on  the  other. 

Draw  the  segment  BC  forming  the  triangle  ABC. 

As  in  Ex.  5  above,  construct  a  triangle  a'b'c'  so  that 
A'b'  =  AB,  b'c'  =  BC,  A'c'  =  AC. 

Show  that  A  ABC  ^  A  a'b'c'  by  one  of  the  tests,  and 
hence  that  Z  A  =  Z.  A',  being  corresponding  angles  of  con- 
gruent triangles,  §  29. 

In  the  above  construction,  would  it  be  wrong  to  make 
AB  =  AC?     Is  it  necessary/  to  do  so  ? 


22  PLANE  GEOMETRY. 

48.  Peoblem.     To  construct  the  ray  dividing  a  given 
angle  into  two  equal  angles,  that  is,  to  bisect  the  angle. 
Solution.     Given  the  angle  A. 

To  construct  the  ray  bisecting  it. 
On  the  sides  of  the  angle  lay  off  n. 

segments    AB     and     AG    so     that  y^  ^^^ /  /  \m 

AB  =  AC.  '       .  /^n 

A^^ — ^ 

With   B   and    C  as   centers  and  '-S 

with  equal  radii  construct  arcs  m  and  n  meeting  at  I). 

Draw  the  segments  CD,  BD,  and  AD. 

Now  show  that  one  of  the  tests  for  congruence  is 
applicable  to  make  AACD  ^  A  ABD. 

Does  it  follow  that  Z  1  =  Z  2  ?     Why  ? 

49.  EXERCISES. 

1.  Is  it  necessary  in  §  48  to  make  AB  =  ACi  In  this  respect  com- 
pare with  the  construction  in  §  47. 

*  2.  Is  any  restriction  necessary  in  choosing  the  radii  for  the  arcs  m 
and  n?  Is  it  possible  to  so  construct  the  arcs  m  and  n,  still  using 
equal  radii  for  both,  that  the  point  D  shall  not  lie  within  the  angle 
BA C ?    In  that  case  does  the  ray  A D  bisect  ABACi 

3.  By  means  of  §  48  bisect  a  straight  angle.  What  is  the  ray  called 
which  bisects  a  straight  angle?  In  this  case  what  restriction  is 
necessary  on  the  radii  used  for  the  arcs  m  and  n  ? 

4.  By  Ex.  3  construct  a  perpendicular  to  a  line  at  a  given  point  in  it. 

5.  Construct  a  perpendicular  to  a  segment  at  one  end  of  it  without 
prolonging  the  segment  and  without  using  the  square  ruler. 

Suggestion.  Let  AB  he  the'  given  segment.  Construct  a  right 
angle  A'B'C  as  in  Ex.'4.    Then  as  in  §  47  construct  Z^ 5 C=  ZA'B'C. 

50.  Definition.  A  line  which  is  perpendicular  to  a  line- 
segment  at  its  middle  point  is  called  the  perpendicular 
bi'sector  of  the  segment. 


RECTILINEAR  FIGURES.  23 

51.  Peoblem.     To  construct  the  perpendicular  bisec- 
tor of  a  given  line-segment.  \1^ 


Solution.      Let    AB   be   the    given 
segment.  ^4<f 


.^'"3 


-^7^". 


'-% 


4\ 


>5 


As  in  §  45,  locate  two  points,  C 
and  D,  each  of  which  is  equally 
distant  from  A  and  B.  /f 

Draw  the  segment  CD  meeting  AB  in  O. 

Then  CD  is  the  required  perpendicular  bisector  of  AB. 

To  prove  this,  show  that  AACD  ^ABCD. 

Hence  Z3  =  Z4.  (Why?) 

By  what  test  can  it  now  be  shown  that 
AAOC^ABOC? 

Hence  Z  1  =  Z  2.  (Why  ?) 

Therefore  CO  (or  CD)  is  perpendicular  to  AB  (Why?) 
and  also  A0=  OB  (Why?). 

It  has  thus  been  shown  that  CD  is  perpendicular  to  AB 
and  bisects  it,  as  was  required. 

52.  The  steps  proved  in  the  above  argument  are : 

(a')  AACD  ^  A  BCD.  (5)Z3  =  Z4.  (c^  AAOC^ABOC. 
(d)  Z 1  =  Z  2,  and  AO  =  BO. 

Study  this.  outHne  with  care.  What  is  wanted  is  the  last  result  (d). 
Notice  that  (d)  is  obtained  from  (c),  (c)  from  (b),  and  (b)  from  (a). 
Thus  each  step  depends  on  the  one  preceding,  and  would  be  impos- 
sible without  it.  To  understand  clearly  the  order  of  the  steps  in  a 
proof  as  shown  by  such  an  outline  is  of  great  importance  in  master- 
ing it. 

53.  EXERCISES. 

1.  In  the  construction  of  §  51,  is  it  necessary  to  use  the  same  radius 
in  locating  the  points  C  and  D  ? 

2.  Name  the  isosceles  triangles  in  the  figure  §51:  (a)  if  the  same 
radius  is  used  for  locating  C  and  D,  (b)  if  different  radii  are  used. 


24 


PLANE  GEOMETRY. 


54.  Problem.     To  construct  a  perpendicular   to   a 
give7i  straight  line  from  a  given  point  outside  the  line. 


Solution.    Let  /  be  the  given  line,  and  P  the  given  point  outside  it. 

With  P  as  a  center  draw  an  arc  cutting  the  line  I  in  two 
points,  A  and  B. 

With  A  and  B  as  centers,  and  with  equal  radii,  draw  the 
arcs  m  and  n  intersecting  in  C.  Draw  the  line  PC  cutting 
I  in  the  point  D. 

Then  the  line  PC  is  the  perpendicular  sought. 

To  prove  this,  draw  the  segments  PA^  PB^  CA^  CB. 

Complete  the  proof  by  showing  that  PC  is  the  perpen- 
dicular bisector  of  AB^  and  hence  is  perpendicular  to  I 
from  the  point  P. 

55.   Problem.     To   construct   a   triangle  ivhen  two 
sides  and  the  included  angle 
are  given. 

Solution.     Let  h  and  c  be  the  /  ^)/ 

given  sides,  and  A  the  given  angle. 

As  in  §  47,  construct  an  angle 
a'  equal  to  /-A.     On  the  sides  of  Z  A'  lay  off  A'B 
A'c  =  h.      Connect  B  and  C. 

Then  A'bc  is  the  required  triangle.     (Why  ?) 


RECTILINEAR   FIGURES.  25 

56.     Problem.     Coiisti^uct  a  triangle  ivhen  hvo  angles 
and  the  included  side  are  given. 

Solution.  Let  Z  A  and  ZB  he 
the  given  angles,  and  c  the  given 
side. 

Construct  Z  A'  =  Z.A. 

On  one  side  of  Z  a'  lay  off 
A^b'  =  e. 

At  b'  construct  Z  A'b'k 
equal  to  Ab. 

Let  b'k  meet  the  other  side  of  Za'  at  C. 

Then  ^'^'c  is  t'he  required  triangle.     (Why  ?) 

57,  EXERCISES. 

1.  If  in  the  preceding  problem  two  different  triangles  are  con- 
structed, each  having  the  required  properties,  how  will  these  triangles 
be  related?     Why? 

2.  If  in  the  problem  of  §  5.5,  two  different  triangles  are  constructed, 
each  having  the  required  properties,  how  will  these  triangles  be  related? 
Why? 

3.  If  two  triangles  are  constructed  so  that  the  angles  of  one  are 
equal  respectively  to  the  angles  of  the  other,  will  the  triangles  neces- 
sarily be  congruent  ? 

4.  If  two  different  triangles  are  constructed  with  the  same  sides, 
how  will  they  be  related  ?     Why  ? 

5.  Construct  an  equilateral  triangle.  Use  §  37  to  show  that  it  is 
also  equiangular. 

58.  We  have  now  seen  that  the  three  tests  for  the  con- 
gruence of  triangles  are  useful  in  making  indirect  measurer 
ments  of  heights  and  distances  when  direct  measurement 
is  inconvenient  or  impossible,  and  also  in  making  numerous 
geometric  constructions.  It  will  be  found,  as  we  proceed, 
that  these  tests  are  of  increasing  usefulness  and  importance. 


26  PLANE  GEOMETRY. 

THEOREMS   AND    DEMONSTRATIONS. 

59.  A  geometric  proposition  is  a  statement  affirming  cer- 
tain properEtes  of  geometric  figures. 

Thus :  "  Two  points  determine  a  straight  line  "  and  "  The  base 
angles  of  an  isosceles  triangle  are  equal "  are  geometric  propositions. 

A  proposition  is  proved  or  demonstrated  when  it  is  shown 
to  follow  from  other  known  propositions. 

A  the,oxem  is  a  proposition  which  is  to  be  proved.  The 
argument  used  in  establishing  a  theorem  is  called  a  proof. 

60.  In  every  mathematical  science  some  propositions 
must  he  left  unproved^  since  every  proof  depends  upon  other 
propositions  which  in  turn  require  proof.  Propositions 
which  for  this  reason  are  left  unproved  are  called  axioms. 

While  axioms  for  geometry  may  be  chosen  in  many  dif- 
ferent ways,  it  is  customary  to  select  such  simple  propo- 
sitions as  are  evident  on  mere  statement. 

61.  Among  the  axioms  thus  far  used  are  the  following : 

Axioms.  I.  A  figure  may  he  moved  about  in  space 
without  changing  its  shape  or  size.     See  §  26. 

II.  Through  two  points  one  and  only  one  straight  line 
can  be  drawn.     See  §§8,  32. 

III.  The  shortest  distance  between  two  points  is  meas- 
ured along  the  straight  line-segment  connecting  them. 

Thus  one  side  of  a  triangle  is  less  than  the  sum  of  the  other  two. 

IV.  //  each  of  two  figures  is  congruent  to  the  same 
figure,  they  are  congruent  to  each  other.     See  §§  28,  40. 

V.  //  a,  h,  c,  d  are  line-segments  (or  angles)  such 
that  a  =  b  and  c  =  d,  then  a-\-c  =  b  +  d  and  a  —  c  =  b  —  d* 

In  the  latter  case  we  suppose  a'^c,  hy-d.     See  §§10,  39. 


RECTILINEAR  FIGURES.  27 

VI.  //  a  and  b  are  line-segments  (or  angles)  such  that 
a=h,  then  a  xn=b  xn  and  a-^n=h-^n;  and  if  a>h, 
then  a  xn>b  xn  and  a^n>h-hn,  n  being  a  "positive 
integer.     See  §§  10,  39. 

Note.  An  equality  or  an  inequality  may  be  read  from  left  to  right 
or  from  right  to  left.     Thus,  a  >  6  may  also  be  read  6  <  a. 

Other  axioms  are  given  in  §§  82,  96, 119,  and  in  Chapter 
VII. 

Certain  other  simple  propositions  may  be  assumed  at 
present  without  detailed  proof.  These  are  called  prelimi- 
nary theorems. 

PRELIMINARY  THEOREMS. 

62.  Two  distinct  lines  can  meet  in  only  one  point. 

For  if  they  have  two  points  in  common,  then  by  Ax.  II  they  are  the 
same  line. 

63.  All  straight  angles  are  equal.     §  20,  Ex.  1. 

64.  All  right  angles  are  equal.     See  Ax.  VI. 

65.  Every  line-segment  has  one  and  only  one  middle 
point. 

See  §  51,  where  the  middle  point  is  found  by  construction. 

66.  Every  angle  has  one  and  only  one  bisector. 

See  §  48,  where  the  bisector  is  constructed. 

67.  One  and  only  one  perpendicular  can  be  drawn 
to  a  line  through  a  point  whether  that  point  is  on  the  line 
or  not.     See  §  20,  Exs.  4,  5;  §  49,  Ex.  4;  §  54. 

68.  The  sum  of  all  the  angles  about  a  point  in  a 
straight  line  and  on  one  side  of  it  is  two  right  angles. 

69.  The  sum  of  all  the  angles  about  a  point  in  a  plane 
is  four  right  angles. 

In  §§  68,  69  no  side  of  one  angle  is  to  lie  inside  another. 


28 


PLANE  GEOMETRY. 


70.  Definitions.  Two  angles  are  said 
to  be  complementary  if  their  sum  is  one 
right  angle.  Each  is  then  called  the 
complement  of  the  other. 

Thus,  Z  a  and  Z  h  are  complementary  angles. 

Two  angles  are  said  to  be  supple- 
mentary if  their  sum  is  two  right 
angles.  Each  is  then  said  to  be  the 
supplement  of  the  other. 

Thus,  Z  1  and  Z.  2  are  supplementary  angles. 

Two  angles  are  called  vertical 
angles  if  the  sides  of  one  are  pro- 
longations of  the  sides  of  the  other. 

Thus,  Z  1  and  Z  3  are  vertical  angles,  and  also  Z  2  and  Z4. 

71.  '  EXERCISES.  ,x    ( 

1.  What  is  the  complement  of  45°?  the  supplement? 

2.  If  the  supplement  of  an  angle  is  140°,  find  its  complement. 

3.  If  the  complement  of  an  angle  is  21",  find  its  supplement." 

4.  Find  the  supplement  of  the  complement  of  30°.     )  y 

5.  Find  the  angle  whose  supplement  is  five  times  its  complement. 

6.  Find  the  angle  whose  supplement  is  n  times  its  complement. 

7.  Find  an  angle  whose  complement  plus  its  supplement  is  110°. 


d\G 


8.  If  in  the  first  figure  Ah  =  2  Aa,  and  Zc=:Za  +  Z  ft,  find  each 
angle. 

9.  If     in    the    second    figure    Z.h  =  lZ.a,    Zc  =  Za  +  Zb,   and 
Z  d  =  Q  Z  a,  find  each  angle. 

10.   If  in  the  third  figure  Zb=  Ze,  Zc^  Za  +  Zh,  Zd  =  2Zb, 
and  Ze  =\Zd^  find  each  angle. 


RECTILINEAR  FIGURES.  29 

PRELIMINARY    THEOREMS. 

72.  Angles  ivhich  are  comjolemeiits  of  the  same  angle 
or  of  equal  angles  are  equal. 

For  they  are  the  remainders  when  the  given  equal  angles  are  sub- 
tracted from  equal  right  angles.     Ax.  V. 

73.  Ajigles  ivhich  are  suppleme7its  of  the  same  angle 
or  of  equal  angles  are  equal. 

For  they  are  the  remainders  when  the  given  equal  angles  are  sub- 
tracted from  equal  straight  angles. 

74.  Vertical  angles  are  equal.      \ 

They  are  supplements  of  the  same  angle. 

75.  If  tioo  adjacent  angles  are  supplementary,  their 
exterior  sides  are  in  the  same  straight  line. 

For  the  two  angles  together  form  a  straight  angle. 

76.  If  tioo  adjacent  angles  have  their  exterior  sides 
in  the  same  straight  line,  they  are  supplementary. 

For  a  straight  angle  is  equal  to  two  right  angles. 

77.  EXERCISES. 

1.  Prove  that  if  one  of  the  four  angles  formed  by  two  intersecting 
straight  lines  is  a  right  angle,  then  all  are  right  angles. 

2.  Show  that  the   rays    bisecting   two   complementary  adjacent 
angles  form  an  angle  of  45°. 

3.  Find  the  angle  formed  by  the  rays  bisecting  two  supplementary 
adjacent  angles.     Prove. 

4.  Find  the  angle  formed  by  the  rays  bisecting  two  vertical  angles. 
Prove. 

5.  The  sum  of  two  adjacent  angles  is  74°.     Find  the  angle  formed 
by  their  bisectors. 

6.  The  angle  formed  by  the  bisectors  of  two  adjacent  angles  is 
37°  18'.     Find  the  sum  of  the  adjacent  angles. 


30  PLANE  GEOMETRY. 

ON  THE  NATURE  OF  A  DEMONSTRATION. 

78.  A  theorem  consists  of  two  distinct  parts,  hypothesis 
and  conclusion. 

In  a  geometrical  theorem,  the  hypothesis  specifies  cer- 
tain properties  which  the  figures  in  question  are  assumed 
to  possess.  The  conclusion  asserts  that  certain  other  prop- 
erties*belong  to  the  figures  whenever  the  assumed  proper- 
ties are  present. 

The  hypothesis  and  conclusion  are  often  intermingled 
in  a  single  statement,  in  which  case  they  should  be  explic- 
itly separated  before  making  the  proof. 

For  example,  in  the  theorem  of  §  37,  The  angles  opposite  the  equal 
sides  of  an  isosceles  triangle  are  equal,  the  hypothesis  is,  "  Two  sides  of 
a  triangle  are  equal,"  and  the  conclusion  is,  "  The  angles  opposite 
them  are  equal." 

79.  If  the  hypothesis  consists  of  several  parts,  these 
should  be  tabulated  and  then  checked  off  as  the  demon- 
stration proceeds.  If  the  theorem  is  properly  stated,  each 
part  of  the  hypothesis  will  be  used  in  the  proof. 

For  instance,  in  the  theorem  of  §  32,  the  hypothesis  is :  AB  —A'B', 
AC=  A' C%  and  Z  A  =  Z  A';  and  the  conclusiou  is:  A  ABC  ^  A  A'B'Cy. 

It  will  be  found  on  examining  the  proof  that  each  part  of  the 
hypothesis  is  needed  and  used  in  the  course  of  the  demonstration. 

If  the  conclusion  could  be  proved  without  using  every  part  of  the 
hypothesis,  then  the  parts  not  used  should  be  omitted  from  the  hypothe- 
sis in  the  statement  of  the  theorem. 

80.  In  the  proof  of  a  theorem  no  conclusion  should  be 
taken  for  granted  simply  from  the  appearance  of  the  figure. 
Each  step  in  a  proof  should  be  based  upon  a  definition,  an 
axiom,  or  a  theorem  previously  proved. 

It  will  then  follow  that  the  theorem  is  as  certainly  true 
as  are  the  simple,  unproved  propositions  with  which  we 
start,  and  upon  which  our  argument  is  based. 


RECTILINEAR  FIGURES, 


31 


Historical  Note.  The  Egyptians  showed  no  knowledge  of  a 
logical  demonstration,  nor  did  the  Arabians,  who  studied  geometry 
quite  extensively.  The  Greeks  developed  the  process  of  demonstra- 
tion to  a  high  state  of  perfection.  They  were  fully  aware,  moreover, 
that  certain  propositions  must  be  admitted  without  proof  (see  §  60). 
Thus  Aristotle  (384-322  B.C.)  says :  *'  Every  demonstration  must  start 
from  undemonstrable  principles.  Otherwise  the  steps  of  a  demon- 
stration would  be  endless."  Euclid  divided  unproved  propositions 
into  two  classes:  axioms,  or  "common  notions,"  which  are  true  of  all 
things,  such  as,  "  If  things  are  equal  to  the  same  thing  they  are  equal 
to  each  other";  and  postulates,  which  apply  only  to  geometry,  such  as, 
"  Two  points  determine  a  line."  The  best  usage  in  modern  mathe- 
matics is  to  adopt  the  one  word  axiom  for  both  of  these,  as  in  §  60. 

Much  practice  is  needed  in  writing  demonstrations  in 
full  detail.  This  should  be  done  in  the  shortest  possi- 
ble sentences,  usually  giving  a  separate  line  to  each  state- 
ment, followed  by  the  definition,  axiom,  or  theorem  on 
which  it  depends. 

For  this  purpose  the  following  symbols  and  abbrevia- 
tions are  convenient: 


Z,  A,  angle,  angles. 

A,  i^,  triangle,  triangles. 

^  ^  r  parallelogram, 
[^parallelograms. 

□,  [s],  rectangle,  rectangles, 
j  right  angle, 
[right  angles, 
[straight  angle. 


rt.Z,  rt.^. 


St.  Z,  St.  A, 

[straight  angles. 

rt.A,rt.A,|"^^^*^'^^^g^"' 
[right  triangles. 

O,  ©,  circle,  circles. 

''^,  ^,  arc,  arcs. 

=  ,  is  equal,  or  equivalent,  to. 

~,  is  similar  to. 

^,  is  congruent  to. 

>,  is  greater  than. 


<,  is  less  than. 
^,  is  less  than  or  equal  to. 
^,  is  greater  than  or  equal  to. 
II,  parallel,  or  is  parallel  to. 
.     r perpendicular,  or 

[is  perpendicular  to. 
lis,  parallels. 
Js,  perpendiculars. 
.'. ,  therefore  or  hence, 
ax.,  axiom, 
th.,  theorem, 
def.,  definition, 
cor.,  corollary, 
alt.,  alternate, 
ext.,  exterior, 
int.,  interior, 
hyp.,  by  hypothesis. 


32  PLANE  GEOMETIiY, 

INEQUALITIES  OF  PARTS  OF  TRIANGLES. 

81.  Definition.  If  one  side  of  a  triangle 
is  produced,  the  angle  thus  formed  is  called 
an  exterior  angle  of  the  triangle. 

Thus,  Z 1  is  an  exterior  angle  of  the  triangle  ABC.   A 

82.  Axiom  VII.  If  a,  h,  c  are  line-segments  {or 
angles)  such  that  a>h  and  h^c,or  such  that  a^h  and 
h>c,  then  a  >  c. 

The  proof  of  the  following  theorem  is  shown  in  full 
detail  as  it  should  be  written  by  the  pupil  or  given  orally, 
except  that  the  numbers  of  paragraphs  should  not  be 
required. 

83.  Theorem.  An  exterior  angle  of  a  triangle  is 
greater  than  either  of  the  opposite  interior  angles. 


Given  the  A  ABC  with  the  exterior  angle  DBC  formed  by  pro- 
ducing the  side  AB. 

To  prove  that  Z.  DBG  >  Z  c  and  also  Z  dbc  >  Z  a. 
Proof:   Let  E  be  the  middle  point  of  BC. 
Find  E  by  the  construction  for  bisecting  a  line-segment  (§  51). 
Draw  AE  and  prolong  it,  making  EF=  AE^  and  draw  BF. 
In  the  two  A  ACE  and  FBE,  we  have  by  construction 
CE  =  EB  and  AE  =  EF. 


RECTILINEAR   FIGURES.  33 

Also  /.  CEA  ==  /.  BEF. 

(Vertical  angles  are  equal,  §  74.) 
.-.  A  ACE  ^  A  FBE, 
(Two  triangles  which  have  two  sides  and  the  included  angle  of  the 
one  equal  respectively  to  two  sides  and  the  included  angle  of  the  other 

are  congruent,  §  32.) 

.-.  /.C=  Z  FBE. 

(Being  angles  opposite  equal  sides  in  congruent  triangles,  §  29.) 

But  ZDBO  Z  FBE. 

(Tf  an  angle  is  the  sum  of  two  angles  it  is  greater  than  either  of 
them,  §39.) 

.-.  Zdbc>  Z  C. 
(Since  Z  DEC  >  Z  FBE  and  Z  FBE  =:ZC,  Ax.  VII,  §  82.)  (%^-[) 

In  order  to  prove  Z  DBC  >  Z  A,  prolong  CB  to  some 
point  G. 

Then  Zabg  =  Zdbc. 

(Vertical  angles  are  equal,  §  74.) 

Now  bisect  AB,  and  in  the  same  manner  as  before  we  may 
prove  ZABG>ZA. 

.'.  Zdbc>  Za. 

(Since  Z  DBC  =  Z  ABG  and  Z  ABG  >ZA,  Ax.  VII,  §  82.) 

For  the  second  part  of  the  proof  let  H  be  the  middle  point 
of  AB.     Draw  CH  and  prolong  it  to  X,  making  cri  =  HK. 

Let  the  student  draw  the  figure  for  the  second  part  of 
the  proof  and  give  it  in  full. 

Hereafter  more  and  more  of  the  details  of  the  proofs  will 
be  left  for  the  student  to  fill  in. 

When  reference  is  made  to  a  paragraph  in  the  text  or 
when  the  reason  for  a  step  is  called  for,  the  complete  state- 
ment of  the  definition,  axiom,  or  theorem  should  be  given 
by  the  student. 


34 


PLANE  GEOMETRY. 


84.  Theokem.  If  two  sides  of  a  'triangle  are  un- 
equal, the  angles  opposite  these  sides,  are  unequal,  the 
greater  angle  being  opposite  the  greater  side. 

Given  ^ABC  in  which  AC>BC. 
To  prove  tliat  Z  ABC  >  Za, 
Proof :    Lay  off  CD  =  CB  and  d  raw  BD. 

Now  give  the  reasons  for  the  follow- 
ing steps:  ^ 


(1)  Z  ABC  >  Z  DBC. 

(2)  ZdBC  =  ZCDB. 

(3)  ZCDB>ZA. 

(4)  .-.  ZABO  ZA. 


(§39) 
(§37) 
(§  83) 
(§82) 


85.  Theorem.  //  two  angles  of  a  triangle  are  un- 
equal, the  sides  opposite  them  are  unequal,  the  greater 
side  being  opposite  the  greater  angle. 

Given  A  ABC  in  which  ZB>ZA. 

To  prove  that  h  >  a. 

Proof:   One  of  the  following  three 
statements  must  be  true: 
(1)  h  =  a,     (2)  h<a,      (3)  h>a. 

But  it  cannot  be  true  that  b  =  a, 
for  in  that  case  Zb  =Za,  (§  37) 

contrary  to  the  hypothesis  that  Z  b  >  Z  A. 

And  it  cannot  be  true  that  h  <  a,  for  in  that  case 

Zb<Za,  (§84) 

contrary  to  the  hypothesis. 
Hence  it  follows  that  b  >  a. 


BECTILINEAB  FIGURES.  35 

86.  The  above  argument  is  called  proof  by  exclusion. 
Its  success  depends  upon  being  able  to  enumerate  all  the 
possible  cases^  and  then  to  exclude  all  but  one  of  them  by 
showing  t^iat  each  in  turn  leads  to  some  contradiction. 

87.  EXERCISES. 

1.  The  hypotenuse  of  a  right  triangle  is  greater  than  either  leg. 

2.  Show  that  not  more  than  two  equal  line-segments  can  be 
drawn  from  a  point  to  a  straight  line. 

Suggestion.     Suppose  a  third  drawn.     Then  apply  §§  37,  83,  84. 

3.  Show  by  joining  the  vertex  A  of  the  triangle  ABC  to  any  point 
of  the  side  BC  that  ZB  +  ZC  <2  rt.  A.     Use  §  83. 

4.  If  tioo  angles  of  a  triangle  are  equal,  the  sides  opposite  them  are 
equal.     Use  §§  84,  86. 

5.  Either  leg  of  an  isosceles  triangle  is  greater  than  half  of  the  base. 
Suggestion.     Use  §  42,  Ex.  4,  and  Ex.  1. 

THEOREMS   ON   PARALLEL   LINES. 

88.  A  straight  line  which  cuts  two  straight 
lines  is  called  a  transversal.  The  various 
angles  formed  ar^  named  as  follows: 

Z4  and  Z5  are  alternate-interior  angles;       ^ 
also  Z  3  and  Z  6. 

Z  2  and  Z  7  are  alternate-exterior  angles ;  also  Z 1  and 
Z8. 

Z 1  and  Z  5  are  corresponding  angles  ;  also  Z  3  and  Z  7, 
Z  2  and  Z  6,  Z  4  and  Z  8. 

Z  3  and  Z  5  are  interior  angles  on  the  same  side  of  the 
trans vei*sal ;  also  Z4  and  Z  6. 

89.  Definition.     Two  complete   lines  which   lie  in  the 
^same  plane  and  which  do  not  meet  are  said  to  be  parallel. 

Two  line-segments  are  parallel  if  they  lie  on  parallel 
lines. 


36  PLANE  GEOMETRY. 

90.  Theorem.  //  two  lines  cut  by  a  transversal  have 
equal  alternate  interior  angles,  the  lines  are  parallel. 

Given  the  lines  /i  and  I2  cut  by  t  so  that 
Z1  =  Z2. 

To  prove  that  Zj  il  l<^.  p^Z 1^ 

Proof :     Suppose    the  lines  l^  and  l^ 
were  to  meet  on  the  right  of  the  trans-  — 
versal.       Then    a    triangle     would    be    / 
formed   of  which   Z  1  is  an   exterior  angle   and  Z  2  an 
opposite  interior  angle. 

This  gives  an  exterior  angle  of  a  triangle  equal  to  an 
opposite  interior  angle,  which  is  impossible.     (Why?) 

Repeat  this  argument,  supposing  l^  and  l^  to  meet  on  the 
left  side  of  the  transversal. 

Hence  Zj  and  Zg  cannot  meet  and  are  parallel  (§  89). 

91.  The  type  of  proof  used  here  is  called  an  indirect 
proof.  It  consists  in  showing  that  something  impossible 
or  contradictory  results  if  the  theorem  is  supposed  not  true. 

92.  Theorem.  //  two  lines  cut  by  a  transversal  have 
equal  corresponding  angles,  the  lines  are  parallel. 

Given  the  lines  /i  and  4  cut  by  t  so  that 
Z  2  =  Z  3.  _ 

To  prove  that  l^  ||  l^. 
Proof  :     Quote  the  authority  for  each 
of  the  following  steps  : 

Z3  =  Z2. 

Z3  =  Z1.  (§74) 


(Ax.  IV) 


.  Z 1  =  Z  2. 

.-.  Z1I1Z2.  (§90) 


RECTILINEAR   FIGURES.  37 

93.  Theorem.  //  two  lines  cut  by  a  transversal  have 
the  sum  of  the  interior  angles  on  one  side  of  the  trans- 
versal equal  to  two  right  angles,  the  lines  are  parallel, 

A 


Given  k  and  l^  cut  by  t  so  that  Z4  +  Z2=2rt.  zi. 

To  prove  that  \  II  l^. 

Proof:     Z  4  is  supplementary  to  Zl  and  also  to  Z  2. 

(Why  ?) 

.-.  Z1  =  Z2.  (Why?) 

.-.  /ill^2-  (Why?) 

94.  EXERCISES. 

1.    Show  that  if  each  of  two  lines  is  perpendicular  to  the  same  line,  they 
are  parallel  to  each  other. 

2.  Let  ABC  be  any  triangle.     Bisect  BC  at  Z).  (^  ^ 
Draw  AD   and    prolong   it   to   make   DE  —  AD.          /NT       \\ 
Draw  CE.     Prove  CE  II  AB.                                            /^/<D 

3.  Use    Ex.   2   to   construct  a   line   through  a    A  ^     J3 
given  point  parallel  to  a  given  line. 

SuGGP]STiON.     Let  AB  be  a  segment  of  the  given  line  and  let  C 
be  the  given  point.     Draw  CA  and  CB  and  proceed  as  in  Ex.  2. 

95.  Exs.  2  and  3  above   show  that  through  a  point  P, 
not  on  a  line  Z,  at  least  one  line  l^  can  be  drawn  parallel  to  I. 

It  seems  reasonable  to  suppose  that  no  other  line  l^  can 
be   drawn  through  P  parallel  to  Z,  al-               „           q 
though  this  cannot  be  proved  from  the           ~X"^Mp^^''^-i^ 
preceding  theorems.     See  §  60.  — '^'^^ ^ 1 

Hence  we  assume  the  following : 


38  PLANE  GEOMETBY. 

96.  Axiom  VIII.  Through  a  point  not  on  a  line  only 
one  straight  line  can  he  drawn  parallel  to  that  line. 

Historical  I^ote.  This  so-called  axiom  of  parallels  has  attracted 
more  attention  than  any  other  proposition  in  geometry.  Until  the 
year  1829  persistent  attempts  were  made  by  the  world's  most  eminent 
mathematicians  to  prove  it  by  means  of  the  other  axioms  of  geometry. 
In  that  year,  however,  a  Russian,  Lobachevsky,  showed  this  to  be  im- 
possible and  hence  it  must  forever  remain  an  axiom  unless  some  other 
equivalent  proposition  is  assumed. 

97.  Theorem.  //  two  'parallel  lines  are  cut  by  a 
transversal,  the  alternate  interior  angles  are  equal. 

Given  /i  II  /g  and  cut  by  t  A 

.  To  prove  that  Z  1  =  Z 2.  rr/^       ^- 

Proof :  Suppose  Z  2  not  equal  to  Z  1.  y  ^ 

Through  P  draw  Zg,  making  Z  3  =  Z 1.    — —/-^ Zi 

.■.\\\\.  (Why?)    / 

But  by  hypothesis  \  H  \  and  thus  we  have  through  P 
two  lines  parallel  to  Z^,  which  is  contrary  to  Ax.  VIII. 

Therefore,  the  supposition  that  Z  2  is  not  equal  to  Z  1 
leads  to  a  contradiction,  and  hence  Z  1  =  Z  2. 

98.  Compare  this  theorem  with  that  of  §  '90.  The  hy- 
pothesis of  either  is  seen  to  be  the  conclusion  of  the  other. 

When  two  theorems  are  thus  related,  each  is  said  to  be 
the  converse  of  the  other.  Other  pairs  of  converse  theo- 
rems thus  far  are  those  in  §  37  and  Ex.  4,  §  87;  §§  75  and 
76,  and  §§  84  and  85. 

The  converse  of  a  theorem  is  never  to  be  taken  for 
granted  without  proof,  sincq  it  does  not  follow  that  a  state- 
ment is  true  because  its  converse  is  true. 

Thus,  it  is  true  that  if  a  triangle  is  equilateral,  it  is  also  isosceles,  but 
the  converse,  if  a  triangle  is  isosceles,  it  is  also  equilateral,  is  not  true. 


-s^- 


OF    THE 


UNIVERSITY 

OF 
.C>^LIFOBg\^^i2J^Cr7L7iV^E^2?   FIGURES,  39 

99.  Theorem.  If  two  'parallel  lines  are  cut  by  a 
transversal,  the  sum  of  the  interior  angles  on  one  side 
of  the  transversal  is  two  right  angles. 

Suggestion.     Make  use  of  the  preced- 
ing theorem  and  give  the  proof  in  full. 
Of  what  theorem  is  this  the  converse  ? 

100.  EXERCISES. 

1.  state  and  prove  the  converse  of  the  theorem  in  §  92. 

2.  Prove  that  if  two  parallel  lines  are  cut  by  a  transversal  the 
alternate  exterior  angles  are  equal.     Draw  the  figure. 

3.  State  and  prove  the  converse  of  the  theorem  in  Ex.  2. 

4.  If  a  straight  line  is  perpendicular  to  one  of  two  parallel  lines,  it 
is  perpendicular  to  the  other  also. 

5.  Two  straight  lines  in  the  same  plane  parallel  to  a  third  line 
are  parallel  to  each  other.     Suppose  they  meet  and  then  use  §  96. 


6/5 


1^11 


D 


6.  If  l^  II  ^2 II  h  and  if  Zl  =  30°,  find  the  other  angles  in  the  first 
figure. 

7.  If  l^Wl^,  how  are  the  bisectors  of  Zl  and  Z3  related?    Of  Z3 
andZ4?   OfZlandZ2?  Use  §  102  for  the  last  case.  3^ 

8.  If  Zj  11/3,  and  ^0=05,  show  that  i)0=OC.                    t^      ^' 
State  this  theorem  fully  and  prove  it.  -  aJ^'^ 1^ 

9.  If /JU2andZ2  =  5Zl,  findZ4andZ3.  ^ 

10.  If  two  parallel  lines  are  cut  by  a  transversal,  the  sum  of  the 
exterior  angles  on  one  side  of  the  transversal  is  two  right  angles. 

11.  State  and  prove  the  converse  of  the  preceding  theorem. 


40  PLANE   GEOMETRY. 

APPLICATIONS  OF   THEOREMS  ON  PARALLELS. 

101.  Problem.  Through  a  given  "point  to  construct  a 
line  parallel  to  a  given  line. 

Given  the  line  /  and  the  point  P  outside  of  it. 

To  construct  a  line  \  through  P  \\  to  I. 

Construction.  Through  P  draw  any  line  making  a  con- 
venient angle,  as  Z 1  with  I.  / 

Through  P  draw  the  line  Zj,  making  , pA        ^ 

Z2  =  Z1(§47).     Then  ?!  II?. 

Proof  :     Use  the  theorem,  §  92.  

Hereafter  all  constructions  should  be        / 
described  fully  as  above,  followed  by  a  proof  that  the  con- 
struction gives  the  required  figure. 

102.  Theorem.  The  sum  of  the  angles  of  a  triangle  is 
equal  to  two  right  angles.  ^ 

Given  A  ABC  with  Z  1,  Z  2,  Z  3.         c  \ 

To  prove  that  \\^         \ 

Zl  +  Z2  +  Z3  =  2rtZs.  \^     ^\4\, 

Proof:   Prolong  ^j5  to  some  point  Z).   ^  B 

Through  B  draw  BE  \\  AC. 

Then  Z5  +  Z4  +  Z3=2rtZs.  (Why?) 

But  Z4  =  Z2and  Z5  =  Z1.  (Why?) 

Hence,  replacing  Z.6  and  Z4  by  their  equals,  Zl  and 
Z2,  wehaveZl  +  Z2  +  Z3  =  2rtZs. 

Historical  Note.  This  is  one  of  the  famous  theorems  of  geome- 
try. It  was  known  by  Pythagoras  (500  B.C.),  but  special  cases  were 
known  much  earlier.  The  figure  used  here  is  the  one  given  by 
Aristotle  and  Euclid.  As  is  apparent,  the  proof  depends  upon  the 
theorem,  §  97,  and  thus  indirectly  upon  Axiom  VIII.  The  interdepend- 
ence of  these  two  propositions  has  been  studied  extensively  during 
the  last  two  centuries. 


RECTILINEAR   FIGURES.  41 

103.  Theorem.     An  exterior  angle  of  a  triangle  is 
equal  to  the  sum  of  the  two  opposite  interior  angles. 

The  proof  is  left  to  the  student. 
Compare  this  theorem  with  that  of  §  83. 

104.  Definition.  A  theorem  which 
follows  very  easily  from  another  theo- 
rem is  called  a  corollary  of  that  theorem. 
U.g.  the  theorem  in  §  103  is  a  corollary 
of  that  in  §  102. 

105.  EXERCISES. 

1.  Find  each  angle  of  an  equiangular  triangle. 

2.  If  one  angle  of  an  equiangular  triangle  is  bisected,  find  all  the 
angles  in  the  two  triangles  thus  formed. 

3.  If  in  SiAABC,AB  =^Cand  ZA  =  Z  5  +  Z  C,  find  each  angle. 

4.  If  in  the  figure  AB  =  AC  and  Zi  =  120°,  find 
Z1,Z2,  Z3. 

5.  If  one  acute  angle  of  a  right  triangle  is  30°, 
what  is  the  other  acute  angle?    If  one  is  41°  23'? 

6.  If  in  two  right  triangles  an  acute  angle  of  one 
is  equal  to  an  acute  angle  of  the  other,  what  can  be 

said  of  the  remaining  acute  angles.     What  axiom  is  involved? 

7.  If  in  two  right  triangles  the  hypotenuse  and  an  acute  angle  of  one 
are  equal  respectively  to  the  hypotenuse  and  an  acute  angle  of  the  other,  the 
triangles  are  congruent.     Prove  in  full. 

8.  Can  a  triangle  have  two  right  angles?  Two  obtuse  angles? 
Can  the  sum  of  two  angles  of  a  triangle  be  two  right  angles?  What 
is  the  sum  of  the  acute  angles  of  a  right  triangle? 

9.  If  two  angles  of  a  triangle  are  given  how  can 
the  third  be  found?  If  the  sum  of  two  angles  of  one 
triangle  is  equal  to  the  sum  of  two  angles  of  another, 
how  do  the  third  angles  compare? 

10.  Prove  the  theorem  of  §  102,  using  each  of  the 
figures  in  the  margin.  The  first  of  these  figures  was 
used  by  Pythagoras. 


42 


PLANE  GEOMETRY. 


106.   Definition.     An  angle  viewed  from 
the  vertex  has  a  right  side  and  a  left  side. 

107-  Theorem.     If  two  angles  have 
their   sides  i^espectively  parallel,   right 
side  to  right  side,  and  left  side  to  left 
side,  the  angles  are  equal. 

Given  Z 1  and  Z  2  such  that  a  II  a'  and  6  li  &' . 

To  prove  that  Z  1  =  Z  2. 

Proof  :    Produce  a  and  b'  till  they  meet, 
forming  Z.  3.     Complete  the  proof. 


Why  do  a  and  h'  meet  when  produced  ? 

Make  a  proof  also  by  producing  h  and  a'  till  they  meet. 

108.  Theorem.  If  two  angles  have  their  sides  re- 
spectively perpendicular,  right  side  to  right  side,  and 
left  side  to  left  side,  the  angles  are  equal. 

Given  Z 1  and  Z  2  such  that  al.a'  and  h  ±  b'- 

To  prove  that  Z1  =  Z2. 

Proof :  Produce  a  and  a^  till  they  meet  in  O. 

Through  o  draw  a  line  parallel  to  h. 

Now  show  that  Z  2  =  Z  3  since  each  is 
the  complement  of  Z4.  Complete  the 
proof. 


109. 


EXERCISES. 


If  two  angles  have  their  sides  respectively 
parallel,  or  perpendicular,  right  side  to  left  side, 
and  left  side  to  right  side,  the  angles  are  supple- 
mentary.    In  each  figure  Zl  =  Z3.  (Why?) 

Historical  Note.  The  theorems  of  §§  107, 
108,  109  are  not  found  in  Euclid's  Elements. 


RECTILINEAR   FIGURES.  43 

OTHER  THEOREMS  ON   TRIANGLES. 

110.  Theoeem.  //  in  two  right  triangles  the  hypote- 
nuse and  one  side  of  one  are  equal  respectively  to  the 
hypotenuse  and  one  side  of  the  other,  the  triangles  are 
congruent,  ^  b'  b 


AiA 


\ 
-'A' 


Given  the  right  A  ABC  and  A'EfC,  having  -45=  A '5**  and 
BC=BfC. 

To  prove  that  A  ABC  ^  A  A'b'c'. 

Proof  :  Place  the  triangles  so  that  j^  and  B^c'  coincide, 
and  so  that  A  and  A'  are  on  opposite  sides  of  BC. 

Then  AC  and  CA'  lie  in  a  straight  line  (Why  ?),  and 
A  aba'  isosceles  (Why?). 

Hence,  show  that  Z  A  =  /.  A'  and  complete  the  proof. 

111.  Corollary.  If  from  a  point  in  a  perpendicular 
to  a  straight  line  equal  oblique  segments  are  drawn  to  the 
line,  these  cut  off  equal  distances  from  the  foot  of  the 
perpendicular,  and  make  equal  angles  with  the  perpendic- 
ular and  with  the  given  line.  ' 

Give  the  proof  in  full,  as  of  an  independent  theorem. 

112.  Theorem.  //  from  a  point  in  a  perpendicular 
to  a  line  oblique  segments  are  drawn,  cutting  off  equal 
distances  from  the  foot  of  the  perpendicular,  these  seg- 
ments are  equal,  and  make  equal  angles  with  the  perpen- 
dicular and  with  the  given  line. 

Give  the  proof  in  full,  using  A  ABA^  of  §  110. 


44 


PLANE  GEOMETRY. 


113.  Theorem.  The  'perpendicular  is  shorter  than 
any  oblique  segment  from  a  point  to  a  line. 

Suggestion.  Show  by  §  102  that  the  angle  opposite  the 
oblique  segment  in  the  triangle  formed  is  greater  than 
either  of  the  other  angles,  and  then  make  use  of  §  85. 

114.  The  distance  from  a  point  to  a  line  means  the 
shortest  distance,  and  hence  is  measured  on  the  perpen- 
dicular. 

115.  Theorem.  If  from  a  point  in  a  perpendicular 
to  a  line  segments  are  drawn  cutting  off  unequal  dis- 
tances from  the  foot  of  the  perpendicular,  the  segments 
are  unequal,  that  segment  heing  the  greater  ivhich  cuts 
off  the  greater  distance. 

Given  AO  ±  BC  and  0C>  OB. 

To  prove  that  AC>  AB. 

Proof  :  Let  ob'  =  OB  and 
draw  AB'. 

Then,  Zob'A<  rt.  Z,  and 
.-.  Z  AB'C  >  rt.  Z. 

Also  Z  OCA  <  rt.  Z. 

.-.aoab'. 

.'.  AC  >  AB  since  AB'  =  AB.    Give  reasons  for  each  step. 

llal6.  Theorem.  If  from  a  point  in  a  perpendicular 
to  a  line  unequal  segments  are  drawn,  these  cut  off  unequal 
distances  from  the  foot  of  the  perpendicidar,  the  greater 
segment  cutting  off  the  greater  distance. 

Suggestion.  Using  the  figure  of  §  115  and  the  hypoth- 
esis that  AC  >  AB,  show  that  two  of  the  following  state- 
ments are  impossible: 

(1)  OC  =  0B  ',   (2)  OC  <  OB  ;  (3)  OC  >  OB.     See  §  86. 


BECTILINEAR  FIGURES. 


45 


117.  Theorem.  //  in  two  triangles  two  sides  of  the 
one  are  equal  respectively  to  two  sides  of  the  other,  hut 
the  included  angle  of  the  first  is  greater  than  the  included 
angle  of  the  second,  then  the  third  side  of  the  first  is 
greater  than  the  third  side  of  the  second. 


Given  A  ABC  and  A'B'C  in  which  AB  =  A'B\  BC  =  B'C'  and 
ZB>ZB'. 

To  prove  that  AOA'c'. 

Proof:  Place  Aa'b'c'  on  A  ABC  so  that  a'b'  coincides 
with  its  equal  AB  and  C'  is  on  the  same  side  of  AB  as  C. 

Let  BD  bisect  Z  c'bc,  meeting  AC  in  D.     Draw  DC' . 

Then  ABDC'  ^ABDC.  (Wby?) 

.\DC'  —  DC  and  AD  -K  DC'  =  AC.         (Why  ?) 

But  ,  AD  +  DC'>AC'.  (Ax.  Ill,  §  61) 

AOAC'.  (Ax.  VII,  §82) 

118.  Theorem.  //  in  two  triangles  two  sides  of  the 
one  are  equal  to  two  sides  of  the  other  hut  the  third  side 
of  the  first  is  greater  than  the  third  side  of  the  second, 
then  the  included  angle  of  the  first  is  greater  than  the 
included  angle  of  the  second. 

Suggestion.  Using  the  figure  of  §  117  and  the  hypothesis 
that  AC>a'c',  show  that  two  of  the  three  following 
statements  are  impossible: 

(1)ZB  =  Zb';  (2)Z5<Zb';  CS')ZB>Zb'.    See  §  86. 


46 


PLANE  GEOMETRY. 


119.  Axiom  IX.  //  a,  h,  c,  d  are  line-segments  (or 
angles)  such  that  a>h  and  c  —  d  or  such  that  a>h  and 
c>d,  then  a-{-  c>h  +  d.  Also  if  a>h  and  cSd,  then 
a  —  c>  h  —  d,  provided  a>c  and  h>d.'  (See  §§  10,  39.) 

120.  Theorem.  The  sum  of  the  segments  drawn  from 
a  point  within  a  triangle  to  the  extremities  of  one  side  is 
less  than  the  sum  of  the  other  two  sides. 

c 


Given  the  point  0  within  the  A  ABC. 
To  prove  that  A0+  OB  <:AC  -\-  CB. 
Proof :  AO  +  OE<AC-\-CE. 

And  OB<OE  +  EB. 

.'.   AO-{-OE-\-OB<AC+ CE+OE+EB. 

Subtracting  OE  from  both  members, 

AO-\-OB<AC  +  CE+EB. 
That  is,  AO  +  OB<AC  +  CB. 


(Why?) 

(Why?) 

(Ax.  IX) 

(Ax.  IX) 


121: 


EXERCISES. 


•^1.    Show  that  any  side  of  a  triangle  is  greater  than  the  difference 
between  the  other  two  sides. 

2.  Show  that  the  sum  of  the  distances  from  any  point 
within  a  triangle  to  the  vertices  is  greater  than  one -half 
the  sum  of  the  sides  of  the  triangle.  -4  B 

3.  Show  that  the  segment  joining  the  vertex  of  an  isosceles  tri- 
angle to  any  point  in  the  base  is  less  than  either  of  the  equal  sides. 

4.  Show  that  any  altitude  of  an  equilateral  triangle  bisects  the 
vertex  angle  from  which  it  is  drawn  and  also  bisects  the  base. 

5.  State  and  prove  the  converse  of  the  theorem  in  Ex.  4. 


RECTILINEAR   FIGURES.  47^ 


6.  Construct  angles  of  60°,  120°,  and  30= 

7.  Construct  angles  of  45°  and  135°. 


Suggestion.     Bisect  a  right  angle  and  extend  one  side. 

122.   Problem.     Given  two  sides  of  a  triangle  and 
an  angle  opposite  one  of  them,  to  construct  the  triangle. 


Solution.    Let  Z  A  and  the  segments  a  and  b  be  the  given  parts. 

On  one  side  oi  Z.  A  lay  off  AB  =  5,  and  let  the  other 
side  be  extended  to  some  point  K. 

With  B  as  a  center  and  a  radius  equal  to  a,  construct 
arcs  of  circles  as  shown  in  the  figure. 

The  following  cases  are  possible : 

(1)  If  a  equals  the  perpendicular  distance  from  B  to 
AK^  the  arc  will  meet  AK  in  but  one  point,  and  a  right 
triangle  is  the  solution. 

(2)  \i  a  <h  and  greater  than  the  perpendicular,  the 
arc  cuts  AK  in  two  points,  and  there  are  then  two  tri- 
angles containing  the  given  parts,  as  shown  by  the  dotted 
lines  in  the  figure. 

(3)  If  a  >  5,  the  arc  will  cut  AK  only  once  on  the 
right  of  J.,  and  hence  only  one  triangle  will  be  found. 

Repeat  this  construction,  making  a  separate  figure  for 
each  case. 

Make  the  construction  when  Z  ^  is  a  right  angle.  Are 
all  three  cases  possible  then?  Make  the  construction 
when  Z  ^  is  obtuse.     What  cases  are  possible  then  ? 


48 


PLANE  GEOMETRY. 


123. 


EXERCISES. 


•-— 1.  A  carpenter  bisects  an  angle  A  as  follows: 
Lay  off  AB  =  AC.    Place  a  steel  square  so  that 

BD  =  CD  as  shown  in  the  figure. 

Draw  the  line  A  D.    Is  this  method 

correct  ?    Give  proof.    Would  this 

method  be  correct  if  the  square 

were  not  right-angled  at  Z)  ? 

2.  In  the  triangle  ABC,  AC  =  BC.  The  points 
D,  E,  F  are  so  placed  that  AD  =  BD  and ^F  =  BE. 
Compare  DE  and  DF.     Prove  your  conclusion. 

3.  If  in  the  figure  Z  2  -  Z 1  =  15°,  and  Z  4  =  120, 
find  each  angle  of  the  triangle. 


-=^  4.   If  in  A^BC,  AC  =  BC,  and  if  ^C  is  extended  to  D  so  that 
AC  =^  CD,  prove  that  DB  is  perpendicular  to  AB. 

5.  In  A  ABC,  CA  =  CB,  AD  =  BE.     Prove  A  ADB  ^  A  ABE. 

6.  In   the   triangle  KLN,    NM   is    perpendicular    to    KL,   and 
KM  -  MN  -  ML.    Prove  that  KLN  is  an  isosceles 
right  triangle. 

-^7.  If  in  the  isosceles  A  ABC  a  point  D  lies  in 
the  base,  and  Z  1  =  Z  2,  determine  whether  there  is 
any  position  for  D  such  that  DE  =  DF. 

8.  If  the  bisectors  AD  and  BE  of  the  base 
angles  of  an  isosceles  triangle  ABC  meet  in  0, 
what  pairs  of  equal  angles  are  formed?  What 
pairs  of  equal  segments ?     Of  congruent  triangles? 

9.  If  the  middle  points  of  the  sides  of  an  equi- 
lateral triangle  are  connected  as  shown  in  the  fig- 
ure, compare  the  resulting  four  triangles. 


BECTILINEAR  FIGURES. 


49 


I 


10.   Triangle  ABC  is  equilateral.     AD  -  BE  =  CF. 
Compare  the  triangles  DBF,  ECF,  FAD. 

AD  B 

11.  Two  railway  tracks  cross  as  indicated  in  the 
figure.  What  angles  are  equal  and  what  pairs  of 
angles  are  supplementary?     State  a  theorem  involved 

in  each  case. 

O 

12.  In  a  A  ABC  does  the  bisector  oi  ZA  also 
bisect  the  side  BC  (])  ii  AC  =  BC  but  AC<AB, 
(2)iiAC  =  AB'i  a 

13.   If  in  the  triangle  yl 5 C,  AB  =  AC  and  if  E 
is  any  point  on  A  C,  find  D  on  AB  so  that 

A  ECB  ^  A  CDB. 

-"M.   If    in    the    figure    AB  =  AC,    find    Zl    if 
ZA  =  60°;  ii  ZA  =  40°.,    Show  that  whatever  the 
^        B  value    of  ZA,  Zl  =  \ZA^rt.Z. 

15.  If  in  the  isosceles  triangle  ABC 
the  middle  point  of  AB  is  D  and 
AF  =z  BE,  find  the  relation  between 
Zl  andZ2. 


16.   In    the    figure    AO  =  OC    and 
0D=  OB.     How   are  the   segments  ^C  and   DB 
related?      How   are   AD   and 
CB  related?    Prove. 


-—17.  To  cut  two  converging 
timbers  by  a  line  AB  which 
shall  make  equal  angles  with 
them,  a  carpenter  proceeds  as 
follows:  Place  two  squares  against  the  tim- 
bers, as  shown  in  the  figure,  so  that  AO  =  BO. 
Show  that  ^^  is  the  required  line. 


50  PLANE  GEOMETRY. 

DETERMINATION  OF  LOCI. 

124.   Theorem.   Every  'point  on   the  bisector  of  an 
angle  is  equidistant  from  the  sides  of  the  angle. 

Given  P,  any  point  on  the  bisector  of  the 
angle  A,  and  PC  and  PB  perpendicular  to 
the  sides. 

To  prove  that  PC  =  PB.  A< 

Proof  :  By  the  hypothesis 

Z1  =  Z2.  c- 

Show  that  ZS  =  Z.4:  and  complete  the  proof. 


125.  Theorem.    //  a  point  is  equally  distant  from  the 
sides  of  an  angle,  it  lies  on  the  bisector  of  the  angle. 

Given  an  angle  A,  any  point  P  and  the  per-  b^ 

pendiculars  PB  and  PC  equal. 

To  prove  that  PA  bisects  the  angle  A. 

Proof :  Give  the  argument  in  full  to 
show  that  A  ABP  ^  A  APC  and  thus 
show  that  Z  1  =  Z  2. 

Hence  AP  is  the  bisector  of  the  angle  A. 

126.  The  two  preceding  theorems  enable  us  to  assert 
the  following : 

(1)  Every  point  in  the  bisector  of  an  angle  is  equidistant 
from  its  sides. 

(2)  Every  point  equidistant  from  the  sides  of  an  angle  lies 
in  its  bisector. 

For  these  reasons  the  bisector  of  an  angle  is  called  the 
locus  of  all  points  equidistant  from  its  sides. 

The  word  locus  means  place  or  position.    It  gives  the  location  of  all 
points  having  a  given  property. 


BECTILINEAR  FIGURES.  51 

127.  All  points  in  a  plane  which  satisfy  some  specified 
condition,  as  in  the  case  preceding,  will  in  general  be 
restricted  to  a  certain  geometric  figure. 

This  figure  is  called  the  locus  of  the  points  satisfying  the 
required  condition,  provided: 

(1)  Every  'point  in  the  figure  possesses  the  required  prop- 
erty, 

(2)  Every  point  in  the  plane  which  possesses  the  required 
property  lies  in  the  figure. 

128.  Theorem.  The  locus  of  all  'points  equidistant 
from  the  extremities  of  a  given  line-segment  is  the  per- 
pendicular bisector  of  the  segment. 

Given  the  perpendicular  bisector  /  meeting  [p 

the  segment  AB  at  i>.  yi  2\ 

To  prove  that  (a)  every  point  in  I  is             / 
equidistant  from  A  and  B;    (5)  every         / 
point  which  is  equidistant  from  A  and  B       / 
lies  in  I.  L 

Proof :   (a)  Let  P  be  any  point  in  I. 
Draw  TA  and  BB. 

Then  BA^BB.  (§112) 

(5)  Let  P  be  any  point  in  the  plane  such  that  BA  —  BB. 

Bisect  Z  ABB  with  the  line  BD. 

Now  show  that  A  ABB  ^  A  BBD. 

And  hence  that  AD  =  BD  and  Z  ADB  =  Z  BDB. 

Hence  BD  is  the  perpendicular  bisector  of  AB^  and  since 
there  is  only  one  such,  this  is  the  line  I  (§  67). 

Thus  the  perpendicular  bisector  of  the  segment  fulfills 
the  two  requirements  for  the  locus  in  question. 

Steps  (a)  and  (5)  together  show  that  a  point  not  on  the 
line  I  is  unequally  distant  from  A  and  B. 


D 


52  PLANE  GEOMETRY. 

129.  Problem.  To  find  the  locus  of  all  points  at  a 
given  distance  from  a  given  straight  line. 

Solution.    Given  the  line  /  and  the  segment  a. 

Construct  a  perpendicular  to  I  at  some  point  A  and  lay 
off  AB  =  a.  ^ 

Through  B  draw  Zj  II  L 

Then  l^  is  a  part  of  the  locus  required.  b        p         ^ 

Proof:   (1)  To  show  that  any  point  P 
in  Zj  is  at  the  distance  a  from  I.  

Draw  PA  and  also  let  fall  PK±l.  a        k 

Now  A  ABP  ^  A  AKP  and  PK  =  AB  =  a.     (Why  ?) 

(2)  To  show  that  any  point  P  in  the  plane  above  I  whose 
distance  from  Z  is  a  lies  in  Zj. 

Draw  a  ±  from  P  to  Z,  meeting  I  and  Zj  in  K  and  p'  re- 
spectively. Then  by  (1)  EP'  =  a.  But  KP  =  a.  Hence 
P  and  P'  coincide  and  P  lies  on  Z^. 

.-.  Zj  is  a  part  of  the  locus  sought. 

Let  the  student  find  another  line  which  is  also  a  part  of 
the  locus. 

Note.  In  (1)  and  (2)  above,  great  care  is  needed  in  keeping  the 
hypothesis  clearly  in  mind. 


130.  EXERCISES. 

1.  Find  the  locus  of  all  points  in  the  plane  equally  distant  from 
two  parallel  lines. 

2.  Find  the  complete  locus  of  all  points  in  the  plane  equally  dis- 
tant from  two  intersecting  lines. 

3.  Find  the  locus  of  all  points  in  the  plane  equally  distant  from  a 
fixed  point. 

4.  Find  the  locus  of  all  points  in  the  plane  equally  distant  from 
two  fixed  points. 


RECTILINEAR   FIGURES. 


53 


^  c 


131.  Theorem.     The  bisectors  of  the  three  angles  of  a 
triangle  meet  in  a  'point 

Given  AD,  BE,  and  CF  bisecting  Z^,  Z5, 
Z  C  respectively  of  the  triangle  ABC. 

To  prove  that  AD^  BE^  and  CF  meet  in 

a  common  point. 

Proof :  Let  AD  and  BE  meet  in  some 

point,  as  O. 

Then  O  is  equidistant  from  AB  and 

AC.     Why? 

Also  O  is  equidistant  from  AB  and  BC.     (Why  ?) 

.-.  O  is  equidistant  from  AC  and  BC.     (Wliy  ?) 

Then  O  lies  on  the  bisector  of  Z  C.     (Why  ?) 

That  is,  CF  passes  through  the  point  o,  and  thus  the 

three  bisectors  meet  in  a  common  point. 

132.  Theorem.     The  three  perpendicular  bisectors  of 
the  sides  of  a  triangle  meet  in  a 
point. 

Given  FH,  DG,  and  EK  perpendicular 
bisectors  of  the  sides  AB^  BC,  and  CA  of 
i^ABC. 

To  prove  that  FH^  DG,  and  EK 
meet  in  a  point. 

Proof:  The  proof  is  exactly  similar  to  that  of  §  131.   0V5  " 

133.  EXERCISES. 

1.  In  §  131  how  do  we  know  that  ^Z)  and  5jE;  meet? 
Suggestion.     Show  that  AD  and  BE  cannot  be  parallel. 

2.  In  §  132  how  do  we  know  that  FH  and  GD  meet? 

3.  Do  the  bisectors  of  the  angles  always  meet  inside  the  triangle  ? 

4.  Do  the  perpendicular  bisectors  of  the  sides  always  meet  inside 
the  triangle?    Draw  figures  to  illustrate  the  various  cases. 


54 


PLANE  GEOMETRY, 


THEOREMS  ON  QUADRILATERALS. 

134.  Definitions.  If  no  three  of  the  points 
-4,  iJ,  C,  D  lie  in  the  same  straight  line,  the 
figure  formed  by  the  four  segments  AB^  BC^ 
CD,  DA,  is  called  a  quadrilateral. 

The  segments  are  the  sides  of  the  quadrilateral  a!iad4he 
points  are  its  vertices. 

Two  sides  are  adjacent  if  they  meet  in  a  vertex,'  as  ^J5  and 
BC.  Otherwise  they  are  opposite,  as  AB  and  CD.  Two 
vertices  are  adjacent  if  they  lie  on  the  same  side,  as  A  and 
B.     Otherwise  they  are  opposite,  as  A  and  C. 

A  diagonal   of   a   quadrilateral   is    a   segment   joining 
two  opposite  vertices,  as  ^C 
and  BD. 

Quadrilaterals  which  have 
a  reentrant   angle,   such    as 
Zbcd,  and  those  in  which 
two  sides  intersect,  such  as  AB  and  CD,  are  not  considered 
here. 


Rhomboid 


Rhombus 


Rectangle 


Square 


135.  A  parallelogram  is  a  quadrilateral  in  which  both 
pairs  of  opposite  sides  are  parallel. 

A  rhomboid  is  a  parallelogram  whose  angles  are  oblique. 
A  rhombus  is  an  equilateral  rhomboid. 

A  rectangle  is  a  parallelogram  whose  angles  are  right 
angles.     A  square  is  an  equilateral  rectangle. 

The  side  on  which  a  parallelogram  is  supposed  to  stand  is 
called  its  lower  base,  and  the  side  opposite  is  its  upper  base. 


RECTILINEAR  FIGURES.  *  55 

136.  A  trapezoid  is  a  quadrilateral  having  only  one  pair 
of  opposite  sides  parallel. 

An  isosceles  trapezoid  is  one  in  which 
the  two  non-parallel  sides  are  equal. 

In  jjj  trapezoid  the  two  parallel  sides 
are  the  upper  and  lower  bases. 

137.  The  altitude  of  a  parallelogram  or  a  trapezoid  is  the 
perpendicular  distance  between  its  bases,  and  its  diameter 
is  the  segment  joining  the  middle  points  of  the  other  sides. 

138.  EXERCISES. 

1.  Name  each  of  the  following  quadrilaterals  on  page  4  :  IHPN^ 
IHGO,  AEDB,  COED,  JMPG,  RSED,  KLWV,  JIBC.  To  deter- 
mine whether  opposite  sides  of  these  figures  are  parallel,  use  the  pro- 
tractor for  measuring  the  necessary  angles. 

2.  Is  every  rectangle  a  parallelogram  ?    Is  the  converse  true  ? 

3.  Is  every  rectangle  a  square  ?    Is  the  converse  true  ? 

139.  Theorem.  Opposite  sides  of  a  parallelogram  are 
equal.  2)  c 


A  B 

Given  ABCD,  a  parallelogram;  that  is,  AB II  CD  and  AD  II BC- 

To  prove  that  AB  =  CD  and  BC=  AD. 

Proof  :    Draw  the  diagonal  AC. 

Prove  A  ABC  ^  A  ACD  and  compare  corresponding  sides. 

What  determines  which  are  corresponding  sides  ? 

140.  EXERCISES. 

1.  Show  that  a  diagonal  of   a  parallelogram  divides  it  into  two 
congruent  triangles. 

2.  Give  the  proof  in  §  139,  using  the  diagonal  BD. 


56  '  PLANE  GEOMETRY. 

141.  Theorem.     The  diagonals  of  a  'parallelogram  bi- 
sect each  other,  n c 


A  B 

Given  O  ABCD  with  its  diagonals  meeting  at  the  point  O. 

To  prove  that  OC  =  OA  and  OB  =  OD. 

Proof :  In  the  triangles  AOB  and  COB^  determine  whether 
sufficient  parts  are  equal  to  make  them  congruent,  and  if 
so  compare  corresponding  parts.  Give  all  the  details  of 
the  proof. 

Can  the  proof  be  given  by  using  the  AAOD  and  BOd  If  so, 
give  it. 

142.  Theorem.  //  a  quadrilateral  has  hath  pairs  of 
opposite  sides  equal,  the  figure  is  a  parallelogram. 

D  '      G 


A  B 

Given  a  quadrilateral  ABCD  in  which  AB  =  CD  and  AD  =  BC. 
To  prove  that  AB  II  CD  and  AD  II  BC. 
Proof  :    Draw  the  diagonal  AC. 

In  the  A  ABC  and  ADC  determine  whether  any  test  for 
congruence  applies,  and  if  so  compare  corresponding  angles. 

143.  EXERCISES. 

1.  By  use  of  the  last  theorem  the  question  of  Ex.  1,  §  138,  can  be 
answered  by  measuring  sides  instead  of  angles.  Verify  the  results 
by  this  process. 

2.  Which  two  of  the  theorems  §§  139,  141,  142  are  converse? 
State  in  detail  the  hypothesis  and  conclusion  of  each. 


RECTILINEAR  FIGURES.  57 

144.  Theorem.  If  a  quadrilateral  has  one  pair  of 
opposite  sides  equal  and  parallel^  the  figure  is  a  parallel- 
ogram. ^  jy 


2"^.. 


A  B 

Given.     (State  all  the  items  given  in  the  hypothesis.) 

To  prove.  (State  what  needs  to  be  proved  in  order  to 
show  that  ABCD  is  a  parallelogram.) 

Proof :    From  the  data  given  prove  that 
A  ABC  ^  A  BBC. 

Use  corresponding  angles  of  these  congruent  triangles  to 
show  that  the  other  two  opposite  sides  are  parallel. 

Hence  show  that  the  figure  is  a  parallelogram. 

Write  out  this  demonstration  in  full. 

Could  the  theorem  be  proved  equally  well  by  drawing 
the  other  diagonal  ?     If  so,  draw  it  and  give  the  proof. 

145.  EXERCISES. 

1.  Prove  that  if  the  diagonals  of  a  quadrilateral  bisect  each  other 
it  is  a  parallelogram.     What  is  the  converse  of  this  proposition  ? 

2.  Show  that  if  two  intersecting  line-segments  bisect  each  othet, 
the  lines  joining  their  extremities  are  parallel. 

3.  The  parallel  lines  Z,  and  l^  are  cut  by  a  trans- 
versal AB.  AC  and  AD  bisect  Z2  and  Zl  respec- 
tively. Prove  that  A  CBA  and  DBA  are  isosceles. 
Compare  the  segments  CB  and  BD. 

4.  If  in  an  isosceles  right  triangle  ABC  the 
bisectors  of  the  acute  angles  meet  at  0,  find  how 
many  degrees  in  the  Z 1  thus  formed. 


58  PLANE  GEOMETRY. 

146.  Theokem.     //  the  two  diagonals  of  a  'parallelo- 
gram are  equal,  the  figure  is  a  rectangle. 


Given  O  ABCD  with  AC  =  BD. 

To  prove  that  Zl  =  Z  2  =  Z3  =  Z4  =  rt.Z. 

Proof :    Show  that  A  ABD  ^  A  ABC. 

.'.  Zl  =  Z3.  (Why?) 

But  Zl  +  ^3  =  2rt.Zs.  (Why?) 

.-.  Zl  =  Z3  =  rt.Z.  (Why?) 

In  like  manner  prove  that  Z  2  =  Z  4  =  rt.  Z. 
Hence  the  figure  is  a  rectangle  (§  135). 

147.  EXERCISES. 

1.  State  and  prove  the  converse  of  the  theorem,  §  146. 

2.  Show  that  a  diameter  of  a  parallelogram  passes  through  the  in- 
tersection of  its  diagonals.     See  §  137. 

Suggestion.     Show  that  it  bisects  each  diagonal. 

3.  Prove  that  the  diagonals  of  a  square  are  perpendicular  to  each 
other. 

4.  Prove  that  the  diagonals  of  a  rhombus  are  perpendicular  to 
each  other. 

5.  Does  the  same  proof  apply  to  Exs.  3  and  4  ? 

6.  Are  the  diagonals  of  a  square  equal?    Is  this  true  of  a  rhom- 
bus?   Prove  each  answer  correct. 

7.  Do  the  diagonals  of  a  square  bisect  each  other?    Is  this  true  of 
a  rhombus?     Of  a  trapezoid ? 

8.  Show  that  if  two  adjacent  angles  of  a  parallelogram  are  equal, 
the  figure  is  a  rectangle. 


BECTILINEAB  FIGURES.  59 

148.  Theorem.  Two  parallelograms  having  an  angle 
and  the  two  adjacent  sides  of  one  equal  respectively  to  an 
angle  and  the  two  adjacent  sides  of  the  other  are  con- 
gruent ^  ^    ^,  ^, 


Given  [U  ABCD  and  A'tlCD'  such  that  AB  =  A'B\  AD  =  A'Df 

andZ^  =  A'. 

To  prove  that  O  ABCD  ^  O  a'b^c'b^. 

Proof:  Apply  O  ABCD  to  CJ  a'b'c'd'  so  as  to  make 
Z.A  coincide  with  Z.A',  AB  falling  on  A'b',  and  AD  on 
A'D'. 

Then  BC  takes  the  direction  b'c\  since  Z.B=/.b\ 
being  supplements  of  the  equal  angles  A  and  A'.    (Why  ?) 

C  falls  on  c\  since  BG  =  jR'o',  being  segments  equal  to 
the  equal  segments  AD  and  A'd'.     (Why  ?) 

Hence  CD  coincides  with  c'd'.     (Why  ?) 

Therefore  O  ABCD  ^  O  a'b'g'd\  since  they  coincide 
throughout. 

149.  EXERCISES. 

1.  Are  two  parallelograms  congruent  if  they  have  a  side  and  two 
adjacent  angles  of  the  one  equal  respectively  to  a  side  and  two  adja- 
cent angles  of  the  other  ?     Draw  figures  to  illustrate  your  answer. 

2.  Are  two  parallelograms  congruent  if  they  have  four  sides  of  one 
equal  to  four  sides  of  the  other?  Show  why,  and  draw  figures  to 
illustrate. 

3.  Compare  the  theorem  of  §  148  and  Exs.  1  and  2  preceding  with 
the  tests  for  congruence  of  triangles. 

4.  Prove  that  the  opposite  angles  of  a  parallelogram  are  equal. 

5.  State  and  prove  the  converse  of  Ex.  4. 


60  PLANE  GEOMETRY. 

OTHER  THEOREMS  APPLYING  PARALLELS 

150.  Theorem.  //  a  line  bisects  one  side  of  a  triangle 
and  is  parallel  to  the  base,  then  it  bisects  the  other  side 
and  the  included  segment  is  equal  c 

to  one  half  the  base. 

Given  a  line  /  il  AB  in   A  ABC  such      y^ — ^ ^» 

that  AD  =^  DC. 

To    prove     that     BE  =  EC    and 
de=Iab. 

Proof  :    Draw  EF  through  E  jjarallel  to  CA. 

Now  show  that  AFED  is  a  parallelogram, 
and  that  A  dec  ^  A  FBE, 

from    which       *  CE  =  EB,  DE  =  FB,  DE  =  AF, 

and  AB  =  AF-{-  FB  =  2de, 

or  DE  =  I  AB. 

State  the  reasons  for  each  step. 

151.  Theorem.  A  segment  connecting  the  middle  points 
of  two  sides  of  a  triangle  is  parallel  to  the  third  side  and 
equal  to  one  half  of  it. 

Given  A  ABC  and  the  segment  DE  such  that  AD  -  DC  and 
CE  =  EB.     See  figure  in  §  1.50. 

To  prove  that  DE  II  AB  and  DE  —  ^  AB. 

Proof :  Since  each  of  the  sides  AC  and  BC  has  but  one 
middle  point  (§  Q>k>).,  it  follows  that  there  is  but  one  seg- 
ment DE  bisecting  both  these  sides.  But  by  §  150  a  certain 
segment  parallel  to  the  base  fulfills  this  condition. 

Hence,  DE  is  parallel  to  the  base  AB. 

Then,  DE  =  1  AB  as  in  §  150. 


RECTILINEAR  FIGURES.  61 

152.  Theorem.  If  a  segment  is  parallel  to  the  bases  of 
a  trapezoid  and  bisects  one  of  the  non-parallel  sides,  then 
it  bisects  the  other  also  and  is  equal  to  one  half  the  sum 

of  the  bases.  d ^c 

eZ. 


A  B 

Given  the  trapezoid  ABCD  in  which  AE  =  ED  and  EFllAB. 
To  prove  that  BF=  FC  and  EF=  ^(AB  +  DC}. 
Proof  :    Draw  the  diagonal  AC  meeting  EF  in  O. 
In  A  ACD,AO  =  OC  and  E0  =  ^ DC.  (Why?) 

In  A  ABC,BF  =  FC  and  OF^^AB.  (Why  ?) 

Adding,  EO+OF  =  EF=  |(^Z?  +  DC).  (§  61,  V) 

Prove  this  theorem  also  by  drawing  the  diagonal  BD. 
State  a  similar  theorem  for  a  parallelogram  and  make  a 
proof  for  that  case  which  is  simpler  than  the  above. 

153.  Theorem.  //  a  segment  connects  the  middle  points 
of  the  two  non-parallel  sides  of  a  trapezoid,  it  is  parallel 
to  the  bases  and  equal  to  one  half  their  sum. 

Proof :  The  argument  is  similar  to  that  in  §  151.  Give 
it  in  full. 

154.  EXERCISES. 

1.  Does  a  theorem  similar  to  tliat  of  §  153  hold  for  a  parallelogram  ? 
If  so,  state  it  and  give  a  simpler  proof  in  this  case. 

2.  If  in  the  figure  DE,  FG,  HI,  etc.,  are  parallel  to  AC  and  if  FEy 
HG,  JI,  etc.,  are  parallel  to  BC,  find  the  sum  of 
BD,  DE,  EF,  FG,  GH,  etc.     How,  if  at  all, 
does  the  length  of  AB  enter  into  the  solution? 

Suggestion.  Produce  GF,  IH,  KJ,  to 
meet  BC,  forming  ZI7.  Then  show  that 
BD  +  EF  +  GH,  etc.,  ^  BC,  and  similarly 
ED+  GF+  IH,  etc.,  =  AC. 


62  PLANE  GEOMETRY. 

155.  Theorem.  //  a  series  of  ^parallel  lines  intercept 
equal  segments  on  one  transversal,  they  intercept  equal 
segments  on  every  transversal. 


Given  the  parallel  lines  k,  Uj  /g,  4,  cutting  the  transversal  ti 
soth&.tAB  =  BC=CD. 

To  prove  that  on  the  transversal  t^  EF=  FG  =  GH. 

Proof :  The  figure  ACGE  is  a  parallelogram  or  a  trapezoid 
according  as  t^  II  t^  or  not. 

in  either  case  BF.,  which  bisects  AC,  also  bisects  EG 
(§152).  .■.EF=FG. 

Similarly,  in  the  figure  DHFB,  FG  =  GH. 
.'.  EF=  FG=  GH. 

156.  Theorem.      The  three  altitudes  of  a  triangle  meet 
in  a  point,  n     tA' 


Y  . 

Outline  of  Proof:  Through  each  vertex  of  the  given 
triangle  ABC  draw  a  line  parallel  to  the  opposite  side, 
forming  a  triangle  A'b'c'. 


RECTILINEAR   FIGURES.  63 

Show  that  ACA'b^  and  AB'cb  are  parallelograms,  and 
hence  that  b'c  =  AB  =  CA'.  That  is,  C  is  the  middle 
point  of  A'b'. 

In  the  same  manner  show  that  A  and  B  are  the  middle 
points  of  b'c'  and  a'c'  respectively.  Also  show  that  the 
altitudes  of  A  ABC  are  the  perpendicular  bisectors  of  the 
sides  of  A  a' b'c',  and  therefore  meet  in  a  point  (§132). 

157.  Definition.  A  segment  connecting  a  vertex  of  a 
triangle  with  the  middle  point  of  the  opposite  side  is  called 
a  median  of  the  triangle. 

158.  Theorem.  The  three  medians  of  a  triangle  meet 
in  a  point  which  is  two  thirds  the  distance  from  each  ver- 
tex to  the  middle  point  of  its  opposite  side. 


A  H  B 

Given  A  ABC  with  medians  BD  and  AE  meeting  in  O. 

To  prove  that  the  median  from  C  also  passes  through  o, 
and  that  AO  =  ^AE,  BO  =  ^  BD,  and  CO  =  |  CH. 

Outline  of  Proof :  Taking  F  and  G?,  the  middle  points  of 
OB  and  OA  respectively,  use  §§  151  and  144  to  show  that 
the  figure  GFED  is  a  parallelogram,  and  hence  that 

1)0=  OF  =  FB  and  EO  =  0G=  GA. 

That  is,  O  trisects  AE  and  BD. 

In  the  same  way  we  find  that  AE  and  CH  meet  in  a 
point  o'  which  trisects  each  of  them. 

Hence  o  and  o'  are  the  same  point.  Therefore  the 
three  medians  meet  in  a  point  which  trisects  each  of  them. 


64  PLANE  GEOMETRY. 

159.  EXERCISES. 

1.  If  one  angle  of  a  parallelogram  is  120°,  how  many  degrees  in 
each  of  the  other  angles  ? 

2.  If  the  angles  adjacent  to  one  base  of  a  trapezoid  are  equal, 
then  those  adjacent  to  the  other  base  are  equal. 

3.  If  the  angles  adjacent  to  either  base  of  a  trapezoid  are  equal, 
then  the  non-parallel  sides  are  equal  and  the  trapezoid  is  isosceles. 

4.  In  an  isosceles  trapezoid  the  angles  adjacent  to  either  base  are 
equal. 

5.  Divide  a  segment  AB  into  three  equal  parts.  ^  ^^^ 
Suggestion.     From  A  draw  a  segment  AC^            ^ 

and  on  it  lay  off  three  equal  segments,  AD^  DE,   j^- 
EF  (§  33). 

JDraw  FB  and  construct  EG  and  DH  each  parallel  to  FB.  Prove 
AH  =  HG  =  GB. 

6.  To  cut  braces  for  a  roof,  as  shown  in  the  figure, 
a  carpenter  needs  to  know  the  angle  DBC  when  the 
angle  DAB  is  given,  it  being  given  that  AB  =  AD. 
Show  how  to  find  this  angle.     (See  §  123,  Ex.  14.) 

7.  If  each  of  the  perpendicular  bisectors  of  the  ^ 

sides  of  a  triangle  passes  through  the  opposite  vertex,  what  kind  of  a 
triangle  is  it  ?  If  it  is  given  that  two  of  the  perpendicular  bisectors  of 
sides  pass  through  the  opposite  vertices,  what  kind  of  a  triangle  is  it? 
If  only  one? 

8.  Find  the  locus  of  the  middle  points  of  the  segments  joining  a 
vertex  of  a  triangle  to  all  points  on  the  opposite  side. 

9.  Given  a  line  I  and  a  point  P  not  in  the  line.  Find  the  locus  of 
the  middle  points  of  all  segments  drawn  from  P  to  l. 

10.  The  length  of  the  sides  of  a  triangle  are  12,  14,  16.  Four  new 
triangles  are  formed  by  connecting  middle  points  of  >  the  sides  of  this 
triangle.     What  is  the  sum  of  the  sides  of  these  four  triangles  ? 

11.  Draw  any  segment  PA  meeting  a  line  Z  in  ^. 
Lay  off  AB  on  I.  With  P  and  B  as  centers  and  with 
AB  and  ylP  as  radii  respectively,  strike  arcs  meet- 
ing in  C.     Draw  PC,  and  prove  it  parallel  to  l. 


RECTILINEAR  FIGURES. 


65 


12.  If  the  side  of  a  triangle  is  bisected  by  the  perpendicular  upon 
it  from  the  opposite  vertex,  the  triangle  is  isosceles. 

13.  State  and  prove  the  converse  of  this  theorem. 

14.  If  in  a  right  triangle  the  hypotenuse  is  twice  as 
long  as  one  side,  then  one  acute  angle  is  60°  and  the 
other  30°. 

Suggestion.  Let  D  be  the  middle  point  of  AB. 
Use  Ex.  12  and  the  hypothesis  to  show  that  A  A  CD  is 
equilateral. 

Prove  the  converse  by  drawing  CD  so  as  to  make  ZBCD  =  ZB. 

15.  A  stairway  leading  from  a  floor  to  one  12  feet  above  it  is  con- 
structed with  steps  8  inches  high  and  10  inches         ^^  .^ 

wide.  What  is  the  length  of  the  carpet  re- 
quired to  cover  the  stairway,  allowing  10  inches 
for  the  last  step,  which  is  on  a  level  with  the 
upper  floor. 

16.  A  stairway  inclined  45°  to  the  horizon- 
tal  leads  to  a  floor  15  feet   above  the  first. 
What  is  the  length  of  the  carpet  required  to 
cover  it  if  each  step  is  10  inches  high  ?    If  each 
is  12  inches?    If  each  is  9  inches? 


Can  this  problem  be  solved  without  know- 
ing the  height  of  the  steps?    Is  it  necessary  to     ^i 
know  that  the  steps  are  of  the  same  height? 


17.  If  the  vertex  angle  of  an  isosceles  triangle  is  60°,  show  that 
it  is  equilateral. 

18.  By  successively  constructing  angles  of  60°  divide  the  perigon 
about  a  point  0  into  six  equal  angles.  (This  is  possible  because 
360  ^  6  =  60.)  With  0  as  a  center  construct  a  circle  cutting  the  sides 
of  these  angles  in  points  A,  B,  C,  D,  F,  F.  Draw  the  segments  AB, 
BC,  etc.  Show  by  Ex.  17  that  each  of  the  six  triangles  thus  formed 
is  equilateral.  Show  also  that  ABCDEF  is  equiangular  and  equi- 
lateral, that  is,  AB  =  BC,  etc.,  and  Z  ABC  =  Z  BCD,  etc. 


66 


PLANE  GEOMETRY. 


POLYGONS. 

160.  Definitions.  A  polygon  is  a  figure  formed  by  a 
series  of  segments,  AB,  BC^  CD,  etc.,  leading  back  to  the 
starting  point  A. 

The  segments  are  the  sides  of  the  polygon  and  the  points 
A,  B,  C,  D,  etc.,  are  its  vertices.  The  angles  A,  B,  C,  Z), 
etc.,  are  the  angles  of  the  polygon. 


A 


E 
Convex. 


F 
Concave. 


Equiangular.        Equilateral.     Regular. 


A  polygon  is  convex  if  no  side  when  produced  enters  it. 
Otherwise  it  is  concave. 

Only  convex  polygons  are  here  considered. 

A  polygon  is  equiangular  if  all  its  angles  are  equal  and 
equilateral  if  all  its  sides  are  equal. 

A  polygon  is  regular  if  it  is  both  equiangular  and  equi- 
lateral. 

A  segment  connecting  two  non-adjacent  vertices  is  a 
diagonal  of  the  polygon. 

The  perimeter  of  a  polygon  is  the  sum  of  its  sides. 

161.  Theorem.  The  sum  of  the  angles  of  a  'polygon 
having  n  sides  is  (2  n  —  4)  right  angles. 

Proof :  Connect  one  vertex  with  each 
of  the  other  non-adjacent  vertices,  thus 
forming  a  set  of  triangles.  Evidently 
the  sum  of  the  angles  of  these  triangles 
equals  the  sum  of  the  angles  of  the 
polygon. 


RECTILINEAR  FIGURES. 


67 


Now  show  that  if  the  polygon  has  n  sides  there  are 
(w  —  2)  triangles.  The  sum  of  the  angles  of  one  triangle 
is  2  rt.  A.  Hence,  the  sum  of  the  angles  of  all  the  tri- 
angles, that  is,  the  sum  of  the  angles  of  the  polygon,  is 

2(7i-2)  =  (2n-4)  rt.  A 

162.  Theorem.  The  sum  of  the  exterior  angles  of  a 
polygon,  formed  by  producing  the  sides  in  succession,  is 
four  right  angles. 


Outline  of  Proof :    The  sum  of  both  exterior  and  interior 
angles  is  2  n  rt.  A.     (Why?) 

The  sum  of  the  interior  angles  is  (2  w  —  4)  rt.  A.  (Why  ?) 
Hence,  the  sum  of  the  exterior  angles  is  4  rt.  A.  (Why  ?) 
Write  out  the  proof  in  detail,  using  the  figure. 


163. 


EXERCISES. 


1.  What  is  the  sum  of  the  angles  of  a  polygon  of  3  sides?  of  4 
sides?  of  5  sides?  of  6 sides?  of  10  sides?  of  18  sides? 

2.  Find  each  angle  of  a  regular  polygon  of  3  sides,  4  sides,  5  sides, 
6  sides,  8  sides,  14  sides,  n  sides. 

3.  Construct  a  regular  triangle,  thus  obtaining  an  angle  of  60". 

4.  Construct  a  regular  quadrilateral.     What  is  its  common  name? 

5.  Prove  that  a  regular  hexagon  ABODE F  may  be  constructed  as 
follows :  Let  A  be  any  point  on  a  circle  with  center  0.  With  A  as 
center  and  OA  as  radius  describe  arcs  meeting  the  circle  in  B  and  in 
F.  With  B  as  center  and  the  same  radius  describe  an  arc  meeting 
the  circle  in  C,  and  so  for  points  to  D  and  E.     See  §  159,  Ex.  18. 


68 


PLANE  GEOMETRY. 


SYMMETRY. 

164.  Two  points  A  and  A'  are  said  to  be   symmetric 
points  with  respect  to  a  line  Z  if  Z  is  the  per- 
pendicular bisector  of  the  segment  AA'. 

A  figure  is  symmetric  with  respect  to  an 
axis  I  if  for  every  point  P  in  the  figure 
there  is  also  a  point  P'  in  the  figure  such 
that  P  and  P^  are  symmetric  points  with 
respect  to  I.  This  is  called  axial  symmetry. 
Two  separate  figures  may  have  an  axis  of 
symmetry  between  them. 

165.  Theorem.     Two  figures  which   are  symmetric 
with  respect  to  a  line  are  congruent. 

Given  two  figures  F  and  F'  symmetric 
with  respect  to  a  line  /. 

To  Prove  that  f^f'. 

Proof:    This  is  evident,  since,  by 
folding  figure  F  over  on  the  line  I 
as  an  axis,  every  point  in  F  will  fall  upon  a  corresponding 
point  in  f'  (Why  ?). 

166.  Corollary.  //  points  A  and  A^ 
and  also  B  and  B'  are  symmetric  with  re- 
spect to  a  line  I,  then  the  segments  AB 
and  A^B^  are  symmetric  with  respect  to  I. 


167.  EXERCISES. 

1.  How  many  axes  of  symmetry  has  a  square ?    A  rectangle?     A 
rhombus?     An  isosceles  trapezoid? 

2.  If  a  diagonal  of  a  rectangle  is  an  axis  of  symmetry,  what  kind 
of  a  rectangle  is  it? 


RECTILINEAR   FIGURES. 


69 


3.  If  a  triangle  has  an  axis  of  symmetry,  what  kind  of 
a  triangle  is  it?  Assume  that  the  axis  passes  through 
one  vertex. 

4.  If  a  triangle  has  two  axes  of  symmetry,  what  kind 
of  a  triangle  is  it  ? 

5.  How  many  axes  of  symmetry  has  an  equilateral  triangle? 

6.  How  many  axes  of   symmetry   has   a 
regular  pentagon  (five-sided  figure)  ? 

7.  How   many  axes  of  symmetry  has   a 
regular  hexagon  ? 

8.  Show  that  one   figure  of   §  40  has  an 
axis  of  symmetry.     State  this  as  a  theorem. 

168.  Problem.  Given  a  polygon  P  and  a  line  I  not 
meeting  it,  to  construct  a  'polygon  P'  such  that  P  and 
P'  shall  he  symmetric  with  respect  to  I. 


Solution.  Let  A,  B,  C,  D,  E  be  the  vertices  of  the  given 
polygon  P,  and  /  the  given  line. 

Construct  A' ^  b',  c\  1>',  e'  symmetric  respectively  to  J., 
J?,  C,  7),  E  with  respect  to  the  line  I. 

Then  the  polygon  P'  formed  by  joining  the  points  A' ., 
b\  c',  d',  e\  a'  in  succession  is  symmetric  to  P. 

Proof :     Give  the  proof  in  full. 

Figures  having  an  axis  of  symmetry  are  very  common 
in  all  kinds  of  decoration  and  architectural  construction. 


70 


PLANE  GEOMETRY. 


169.    Two  points  A  and  A'  are  symmetric  with  respect 
to  a  point  O  if  the  segment  A  A'  is  bisected  by  o. 

A  figure  is  synimetric  with  respect  to  a 
point  O  if  for  every  point  P  of   the  figure 
there  is  a  point  p'  also  in  the  figure  such  that  P  and  p' 
are  symmetric  with  respect  to  O. 

Such  a  figure  is  said  to  have  central  sym-  ^^ 

metry  with  respect  to  the  point.     The  point    /\;;--.(0// 
is  called  the  center  of  symmetry.     A  circle    / /^'''    ^-^ 
has  central  symmetry.  '^^ 

Two  separate  figures  may  have  a  center  of  symmetry 
between  them. 


170. 


EXERCISES. 


1.  Prove  that  if  A  and  A'  and  also  B  and  B'  are  symmetric  with 
respect  to  a  point  0,  then  the  segments  AB  and  A'B'  are  symmetric 
with  respect  to  0. 

2.  Prove  that  if  the  triangles  ABC  and  A'B'C  are  symmetric  with 
respect  to  a  point  0,  then  they  are  congruent. 

171.  Problem.  Given  a  polygon  P  and  a  point  0 
outside  of  it,  to  construct  a  polygon  P'  symmetric  to  P 
with  respect  to  0. 


Solution".  Construct  points  symmetric  to  the  vertices 
of  P.  Connect  these  points,  forming  the  polygon  P',  and 
prove  this  is  the  polygon  sought. 


D' 


\P 


^V 


D" 


BECTILINEAR  FIGURES.  .        71 

172.  Theorem.  //  a  figure  has  two  axes  of  symmetry 
at  right  angles  to  each  other,  their  point  of  intersection 
is  a  center  of  symmetry  of  the  figure. 

Outline  of  Proof :  It  is  to  be  shown  that  for  every  point 
P  in  the  figure,  a  point  p"  also  in  the  figure  can  be  found 
such  that  PO  =  P"o  and  POP"  is  a 
straight  line.  Draw  PP'  ±  Z,  p'p" 
±  V,  p"p"'  ±  I  and  connect  the 
points  P'"  and  P. 

Now  use  the  hypothesis  that  I LV, 
and  each  an  axis  of  symmetry,  to 
show  that  pp'pf'p'"  is  a  rectangle  ^" 
of  which  DD"  and  b'd"'  are  the 
diameters.  Hence  O,  the  inter- 
section of  DD"  and  d'd'",  bisects  the  diagonal  PP",  making 
p"  symmetric  to  P  with  respect  to  O,  §  147,  Ex.  2.  Give 
the  proof  in  full. 

173.  EXERCISES. 

1.  Has  a  square  a  center  of  symmetry?  has  a  rectangle? 

2.  If  a  parallelogram  has  a  center  of  symmetry,  does  it  follow  that 
it  is  a  rectangle  ? 

.  3.   Has   a  trapezoid   a   center    of    symmetry?    has    an    isosceles 
trapezoid  ? 

4.  If  two  non-parallel  straight  lines  are  symmetric  with  respect 
to  a  line  I,  show  that  they  meet  this  line  in  the  same  point  and  make 
equal  angles  with  it.  (Any  point  on  the  axis  of  symmetry  is  regarded 
as  being  symmetric  to  itself  with  respect  to  the  axis.) 

5.  If  segments  AB  and  A'B'  are  symmetric  with  respect  to  a  point 
0,  they  are  equal  and  parallel. 

6.  Has  a  regular  pentagon  a  center  of  symmetry  ?  See  figure  of 
Ex.  6,  §  167. 

7.  Has  a  regular  hexagon  a  center  of  symmetry  ?     Ex.  7,  §  167. 

8.  Has  all  equilateral  triangle  a  center  of  symmetry  ? 


72         .  PLANE  GEOMETRY. 

METHODS  OF  ATTACK. 

174.  No  general  rule  can  be  given  for  proving  theorems 
or  for  solving  -problems. 

In  the  case  of  theorems  the  following  suggestions  may 
be  helpful. 

(1)  Distinguish  carefully  the  items  of  the  hypothesis  and 
of  the  conclusion. 

It  is  best  to  tabulate  these  as  suggested  in  §  79. 

(2)  Construct  with  care  the  figure  described  in  the  hy- 
pothesis. 

The  figure  should  be  as  general  as  the  terms  of  the  hypothesis  per- 
mit. Thus  if  a  triangle  is  called  for  but  no  special  triangle  is  men- 
tioned, then  a  scalene  triangle  should  be  drawn.  Otherwise  some 
particular  form  or  appearance  of  the  figure  may  lead  to  unwarranted 
conclusions. 

(3)  Study  the  hypothesis  ivith  car,e  and  determine  whether 
any  auxiliary  lines  may  assist  in  deducing  the  properties 
required  by  the  conclusion. 

Study  the  theorems  previously  proved  in  this  respect.  A  careful 
review  of  these  proofs  will  lead  to  some  insight  as  to  how  they  were 
evolved. 

175.  Direct  Proof.  The  majority  of  theorems  are  proved 
by  passing  directly  from  the  hypothesis  to  the  conclusion 
by  a  series  of  logical  steps.     This  is  called  direct  proof. 

It  is  often  helpful  in  discovering  a  direct  proof  to  trace 
it  backward  from  the  conclusion. 

Thus,  we  may  observe  that  the  conclusion  C  follows  if  statement 
B  is  true,  and  that  B  follows  if  A  is  true.  If  then  we  can  show  that  A 
is  true,  it  follows  that  B  and  C  are  true  and  the  theorem  is  proved. 

Having  thus  discovered  a  proof,  we  may  then  start  from 
the  beginning  and  follow  it  directly  through. 
As  an  example  consider  the  following  theorem: 


RECTILINEAR  FIGURES.  73 

The  segments  connecting  the  middle  points  of  the  opposite 

sides   of  any   quadrilateral    bisect  P 

each  other. 

s< 

Given  segments  AB  and  CD  connect- 
ing the  middle  points  of  the  quadrilat- 
eral PQRS. 

To  prove  that  AB  and  CD  bisect  each  other. 

Proof :  Draw  the  diagonals  PR  and  SQ  and  the  segments  AD,  DB, 
BC,  CA. 

Xow  AB  and  CD  bisect  each  other  if  ADBC  is  aO,  and  ADBC  is 
a  Oil  AD  W  CB  and  A  C  II  DB. 

But  AD  II  CB  since  each  is  II  SQ.     See  §  151. 

And  A  C  II  DB  since  each  is  II  PR. 

Hence  ADBC  is  aOand  AB  and  CD  bisect  each  other. 

Notice  that  the  auxiliary  lines  PR  and  SQ  divide  the 
figure  into  triangles  and  this  suggests  the  use  of  §  151. 

176.  Indirect  Proof.  In  case  a  direct  proof  is  not  easily 
found,  it  is  often  possible  to  make  a  proof  by  assuming 
that  the  theorem  is  not  true  and  showing  that  this  leads  to 
a  conclusion  known  to  he  false. 

As  an  example  consider  the  follow- 
ing theorem  : 

A  convex  polygon  cannot  have  more 
than  three  acute  angles. 

Proof :     Assume    that    such    a    polygon 
may  have  four  acute  angles,  as  Z  1,  Z2,  Z3,  Z4. 

Extend  the  sides  forming  the  exterior  angles  5,  6,  7,  8. 

Since  Z  1  +  Z  5  =  2  rt.  ^,  Z  2  +  Z  6  =  2  rt.  A,  etc.,  and  since  Z 1, 
Z2,  Z  3,  Z  4  are  all  acute  by  hypothesis,  it  follows  that  Z  5,  Z  6,  Z  7,  Z8 
are  all  obtuse,  and  hence  Z  5  4-  Z  6  +  Z 7+  Z 8  >  4  rt.  zi. 

But  this  cannot  be  true  since  the  sum  of  all  the  exterior  angles  is 
exactly  4  rt.  ^  by  §  162. 

Hence  the  assumption  that  Zl,  Z2,  Z3,  Z4  are  all  acute  is  false. 
That  is,  a  convex  polygon  cannot  have  more  than  three  acute  angles. 


74  PLANE  GEOMETRY. 

The  proof  by  the  method  of  exclusion  (§  86)  involves 
the  indirect  process  in  showing  that  all  but  one  of  the  pos- 
sible suppositions  is  false. 

177.  The  solution  of  a  problem  often  involves  the  same 
kind  of  analysis  as  that  suggested  for  the  discovery  of  a 
direct  proof  (§  175). 

For  instance,  consider  the  following  problem : 

To  draw  a  line  parallel  to  the  base  of  a  triangle  such  that 
the  segment  included  hetwee^i  the  sides  shall  equal  the  sum  of 
the  segments  of  the  sides  between  the  parallel  and  the  base. 

Given  the  A  ABC.  c 

To  find  a  point  P  through  which  to  draw  DE  WAB  y\ 

so  that  DP+  PE  =  AD-^r  EB.  pX^P  \?<7 

Construction.     Draw  the  bisectors  oiAA  and  B  ..g^^^V"^       r^ 

meeting  in  point  P.  A.  B 

Then  P  is  the  point  required. 

Proof :  Z 1  =  Z 3  since  DE  WAB,  and  Zl  =  Z2  since  BP  bisects Z B. 

.-.  Z  3  =  Z  2  and  PE  =  BE.     (Why  ?) 

Likewise  in  A  A  DP,  AD  =DP. 

Hence  DP+  PE  =  AD+  BE.     (Ax.  Y.) 

Or  DE  =  AD  +  BE. 

This  construction  is  discovered  by  observing  that  a  point  P  must  be 
found  such  that  PE  =  BE  and  PD  =  AD.  This  will  be  true  if 
Z3=:Z2  and  this  follows  if  Zl  =  Z2,  while  at  the  same  time  Z6  =Z5 
and  Z  4  =  Z  5.  Hence  the  bisectors  of  the  base  angles  will  determine 
the  point  P. 

Having  thus  discovered  the  process,  the  construction  and  proof  are 
made  directly. 

178.  In  general  the  most  effective  help  is  a  ready  knowl- 
edge of  the  facts  of  geometry  already/  discovered,  and  skill 
in  applying  these  will  come  with  practice.  It  is  important 
for  this  purpose  that  summaries  like  the  following  be  made 
by  the  student  and  memorized: 


BECTILINEAR  FIGUBES.  75 

179.    (fl^)  Two  triangles  are  congruent  if  they  have: 

(1)  Two  sides  and  the  included  aiigle  of  the  one  equal  to 
the  corresponding  parts'of  the  other. 

(2)  Two  angles  and  the  included  side  of  the  one  equal  to 
the  corresponding  parts  of  the  other. 

(3)  Three  sides  of  the  one  equal  respectively  to  three  sides 
of  the  other. 

In  the  case  of  right  triangles  : 

(4)  The  hypotenuse  and  one  side  of  one  equal  to  the  corre- 
sponding parts  of  the  other. 

(5)  Two  segments  are  proved  equal  if: 

(1)  They  are  Homologous  sides  of  congrueyit  triangles. 

(2)  They  are  legs  of  an  isosceles  triangle. 

(3)  They  are  opposite  sides  of  a  parallelogram. 

(4)  They  are  radii  of  the  same  circle. 

SUMMARY   OF   CHAPTER  I. 

1.  Make  a  summary  of  ways  in  which  two  angles  may  be  shown  to 
be  equal. 

2.  Make  a  summary  of  ways  in  which  lines  are  proved  parallel. 

3.  What  conditions  are  sufficient  to  prove  that  a  quadrilateral  is  a 
parallelogram  ? 

4.  Make  a  list  of  problems  of  construction  thus  far  given. 

5.  Make  a  list  of  definitions  thus  far  given.     Which  of  the  figures 
defined  are  found  on  page  4? 

6.  Tabulate  all  theorems  on 

(a)  bisectors  of  angles  and  segments, 

(6)  perpendicular  lines, 

(c)  polygons  in  general, 

(d)  symmetry. 

'    7.    What  are  some  of  the  more  important  applications  thus  far 
given  of  the  theorems  in  Chapter  I  ? 


76 


PLANE  GEOMETRY. 


Border,  Parquet 
Flooring. 


PROBLEMS   AND   APPLICATIONS. 

1.  Divide  each  side  of  an  equilateral  triangle 
into  three  equal  parts  (§  159,  5),  and  draw  lines 
through  the  division  points  as  shown  in  the  figure. 

Prove  (a)  The  six  small  triangles  are  equilateral 
and  congruent  to  each  other.  (First  prove  them 
equiangular.) 

(b)  The  two  large  triangles  are  congruent. 

(c)  The  inner  figure  is  a  regular  hexagon. 

2.  Let  Zj  and  h  be  parallel  lines,  with  A  E  per- 
pendicular to  both.  Lay  off  segments  AB,  BC, 
CD,  .  .  .  and  EF,  FG,  GH,  .  .  .  each  equal  to 
AE.     Connect  these  points  as  shown  in  the  figure. 


H 


Tile  Floor  Border. 


Parquet  Floor  Border. 

Prove  (a)  EKA,  EFK,  A  KB,  BLC,  etc.,  are  congruent  right  isosceles 
triangles. 

(h)  FLBK,  etc.,  are  squares. 

Bisect  AB,  BC,  . . .  EF,  FG  . . .  and  join  points  by  lines  parallel 
to  J&5  and  ^irrespectively.  Notice  that  the  resulting  figure  forms 
the  basis  for  the  floor  border  to  the  right. 

3.  Let  Zj  h  be  parallel  lines,  with  MN  perpendicular  to  both. 
Through  F,  the  middle  point  of  MN,  draw  lines  KE  and  A  G,  each 
making  an  angle  of  30°  with  MN. 

Q      H     G  M  E     D 


Parquet  Floor  Border. 

Prove  FE  =  AFhj  showing  /\ANF^/\  FEM. 

Also  prove  KF  =  FG  and  that  ^  FKA  and  FEG  are  equilateral. 


BECTILINEAR  FIGURES. 


77 


Lay  off  segments  AB,  BO,  ED,  DP,  KJ,  etc.,  each  equal  to  AK. 
Connect  points  as  shown  in  the  figure. 

Prove  ABCO^A  CPD  ^  A  AFK,  etc. 

Prove  that  ABCDEF  and  JKFGHl  are  regular  hexagons. 

4.  A  network  of  congruent  equilateral  tri- 
angles is  constructed  in  a  rectangle  ABCD  as 
shown  in  the  figure. 

•     (a)  How  many  of  these  triangles  meet  in  a 
common  vertex? 

(b )  Does  this  n umber  of  equilateral  triangles 
exactly  cover  the  whole  plane  about  the  vertices? 
Why? 

(c)  Do  the  triangles  that  meet  in  one  point 
form  a  regular  hexagon  ? 

(d)  At  what  angle  to  the  horizontal  lines 
are   the  oblique  lines  that  form  sides  of  the 


triangles?  e.g.  what  is  the  angle  DCK'i 


TTTTTl 
rWTTTT 


(e)  Compare   the  lengths  of  CK  and  AK 
(see  §  159,  Ex.  M).  Tile  Flooring. 

To  construct  this  network  when  a  side  a  of 
the  triangles  is  given,  proceed  as  follows :  From  the  vertex  J.  of  a 
right  angle  lay  off  segments  equal  to  a  along  one  side  AK.  With  one 
of  these  division  points  K  as  center  and  with  a  radius  equal  to  twice 
AK  strike  an  arc  meeting  the  side  AC  at  C.  Draw  CD  parallel  to 
AB  and  lay  off  segments  on  it  equal  to  a.  Connect  these  points  as 
shown  in  the  figure,  and  through  the  intersection  points  draw  the 
horizontal  lines,  thus  fixing  the  division  points  on  ^  C  and  BD. 

Prove  that  the  resulting  triangles  are  equilateral  and  congruent  to 
each  other. 

Suggestion.  Use  in  order  the  converse  of  §  159,  Ex.  14 ;  §  144,  and 
the  fact  that  a  diagonal  divides  a  parallelogram  into  two  congruent 
triangles. 

Notice  how  this  construction  is  studied.  The  figure  is  first  sup- 
posed constructed  and  its  properties  tabulated.  Some  of  these  (c)  and 
(d)  are  then  used  in  making  the  construction.  This  method  is  of 
very  general  application  in  problems  of  construction. 

Note.  This  is  the  method  by  which  a  designer  would  construct  a 
network  of  congruent  regular  triangles. 


78 


PLANE  GEOMETRY, 


5.  Construct  a  network  of  triangles  as  in  Ex.  4,  using  pencil  and 
then  ink  in  parts  of  the  lines,  making  a  set  of  regular  hexagons  as 
shown  in  the  figure. 

How  many  such  hexagons  meet  in  a  point?    Will  this  number 
,9  exactly  cover  the  plane  about  the  point?    Why? 


^ 

/  \  f  \  /  \ 

y<y^)i 

■■ 

1 

>-H-H-< 

T"^^T" 

E¥3 


Tile  Flooring. 


6.  Point  out  how  the  figure  constructed  under  Ex.  5  may  be  made 
the  basis  for  the  two  tile  floor  patterns  here  given. 

7.  Given  a  rhombus  whose  acute 
angles  are  60  °.  Show  how  to  cut  off  two 
triangles  so  the  resulting  figure  shall  be 
a  regular  hexagon.  What  fraction  of 
the  whole  rhombus  is  thus  cut  off? 
What  fraction  of  the  pattern  is  black  ? 

8.  In  a  parallelogram,  prove  : 

(«)  The  diameters  divide  it  into  four  congruent  parallelograms. 

{h)  The  segments  joining  the  midpoints  of  the  sides  in  order  form 
a  parallelogram,  and  the  four  triangles  thus  cut  off  may  be  pieced 
together  to  form  another  parallelogram  congruent  to  the  one  first 
formed. 

(c)  The  diameters  and  diagonals  all  meet  in 
a  point. 

Suggestion.  Prove  that  a  diameter  bisects 
a  diagonal.  Do  these  propositions  hold  for  a 
square?  What  part  of  each  of  the  three  large 
squares  in  the  adjoining  figure  is  white? 

9.  The  octagons  (eight-sided  polygons) 
in  the  figure  are  equiangular  and  the  small 
quadrilaterals  are  squares. 
%v^  •  Find  the  angles  of  the  irregular  hexagons. 
If  the  octagons  were  regular,  would  the 
hexagons  be  equiangular  ?  equilateral  ?  reg- 
ular? Tile  Pattern. 


^^"•^'J!li|iiii7'ili:i!!l'yTil 


XT  :^ 


.U. 


Tile  Border. 


BECriLINEA  R  FIG  URES. 


79 


Tile  Pattern. 


10.  In   the   figure   the   octagons    are   equi- 
angular. 

(a)   Will  two  such  octagons  and  a  square 
completely  cover  the  plane  about  a  point  ?  Why  ? 

(J))  At  what  angles  do  the  slanting  sides  of 
the  octagons  meet  the  horizontal  line  ? 

Find  all  the  angles  of  the  two  small  white 
trapezoids  to  the  right. 

(c)  Could  a  pattern  of  this  type  be  constructed  by  using  regular 
octagons  ? 

11.  Equilateral  triangles  and  regular  hexagons  have  thus  far  been 
found  to  make  a  complete  pavement.  What  other  regular  polygons 
can  be  used  to  make  a  complete  pavement? 

'^  12.  If  three  regular  pentagons  (five-sided  polygons)  meet  in  a 
common  vertex,  will  they  completely  cover  the  plane  about  that 
point?    If  not,  by  how  great  an  angle  will  they  fail  to  do  so? 

V  13.  If  four  regular  pentagons  be  placed  about  a  point,  by  how 
great  an  angle  will  they  overlap? 

14.  If  three  regular  seven-sided  polygons  are  placed  about  a  point, 
by  how  great  an  angle  will  they  overlap?     If  two,   by  how  many 

^degrees  will  they  fail  to  cover  the  plane? 

In  Exs.  12,  13,' 14  it  is  understood  that,  so  far  as  possible,  the  poly- 
gons are  so  placed  that  no  part  of  one  of  them  lies  within  another; 
that  is,  they  are  not  to  overlap. 

15.  Answer  questions  like  those  in  Ex.  14  for  regular  polygons  of 
8,  9,  and  10  sides.  Can  any  regular  polygon  of  more  than  six  sides  be 
used  to  form  a  complete  pavement? 

Note  that  the  larger  the  number  of  sides  of  a  regular  polygon  the 
larger  is  each  angle. 

16.  A  carpenter  divides  a  board  into  strips  of  equal  width  as  fol- 
lows :  Suppose  five  strips  are  desired.  Place  a  steel  square  in  two 
positions,  as  indicated  in  the  figure,  at  such 
angles  that  the  distance  in  inches  diagonally 

across  the  board  shall  be  some  multiple  of  five 

(in  the  figure  this  distance  is  15  inches).     Mark  v        ^ 


the  points  and  connect  them  as  shown  in  the  figure, 
lines  divide  the  board  into  equal  strips. 


Prove  that  these 


80 


PLANE  GEOMETRY. 


(The  black  and  white  figures  on  this  page  are  parquet  floor  patterns.) 

17.  Given  an  isosceles  right  triangle  ABC  with  the  altitude  CO 
upon  the  hypotenuse. 

(a)  Show  how  to  draw  xyWAC  such  that  xy  =  Cy. 
Suggestion,     Bisect  Z  OCA  and  let  the  bisector 

meet  AB  in  x. 

Draw  xy  II A  C.     Prove  A  xyC  isosceles. 

(b)  Prove  xyCA  s^u  isosceles  trapezoid. 

(c)  Draw  yz  II  OB  and  prove  yz  =  yx. 

Suggestion.     Prove  A  Cyz  isosceles. 

(d)  Prove  that    xyCA    and   xyzB  are  congruent 
trapezoids. 

18.  ABCD  is  a  square.     Lines  are  drawn  as  shown 
in  the  figure  so  that  Z1  =  Z2  =  Z3  =  ,Z4. 

(a)  Prove  A  ABy  ^  A BCz  ^  A  CDw  ^ADAx. 

(b)  Prove  each  of  these  triangles  a  right  triangle. 

(c)  Prove  that  xyzw  is  a  square. 

(d)  If   Z 1  =  45°,  what  can  be  said  of  the  figure 
xyzw  ? 

This  ^nd  the  following  design  are  of  Arabic 
origin. 

19.  On  the  sides  of  a  square  ABCD  the  points  E,  F,  G,  H  are  laid 
ofe  so  that  AE  =  BF=  CG  =  DH.     Ax  and  zC,  By  and  wD  are  in  the  • 
diagonals  of  the  square  and  Ey,  Fz,  Gw,  and  Hx  are  parallel  to  these.^ 

D  G   C 


A  E 
Prove  that : 

(a)  AxH,  EyB,  etc.,  are  congruent  right  isosceles  triangles. 

(b)  AEyx,  BFzy,  etc.,  are  congruent  parallelograms. 

(c)  xyzta  is  a  square. 

(d)  If  AB  =  a,  find  how  long  A  E  should  be  taken  in  order  that  xy 
shall  equal  EB ;  'also  equal  to  one  half  EB. 


RECTILINEAR   FIGURES. 


81 


G    C 


20.   A  BCD  is  a  square,  AE  =  BF=  CG  =  DH  and  A  G,  CE,  BH, 
and  DF  are  drawn  as  shown  in  the  figure. 
Prove  that : 

(a)  A  G  W  EC  and  BHWFD. 

(b)  Triangles  HwA,  ExB,  etc.,  are 

congruent. 

(c)  The  four  trapezoids  A  Exw,  etc., 

are  congruent. 

(d)  xyzw  is  a  square. 


D      N     M     G 


^m 


Parquet  Flooring. 

21.   In   the   square  ABCD,  AF  =  AG  =  HB  =  BK,  etc.,  and  the 
figure  is* completed  as  shown. 

(a)    Pick  out  all  isosceles  triangles.     Prove. 

(/>)"  Pick  out  all  congruent  triangles.     Prove. 

(c)    Has  this  figure  one  or  more  axes  of  syiiimetry? 


D  M 


L     C 


A  F 


G   B 


Linoleum  Pattern. 


22.    On  the  sides  of  the  square  ABCD  points  E,  F,  G,  H,  .  .  . 
are  taken,  so  that  AE  =  AF  =  GB  =  BH  =  etc. 

The  middle  points  of  EF,  GH,  KL  and  MN  are  connected,  form- 
ing the  quadrilateral  PR  ST. 
Prove  that : 

(a)    AC  \^  the  perpendicular  bisector  of  EF  and  KL. 
(V)    FGRP  is  an  isosceles  trapezoid. 

(c)  PRT  and  RST  are  congruent  right  triangles. 

(d)  PR  ST  is  a  square. 


82 


PLANE  GEOMETRY. 


23.  Ou  the  sides  of  a  square  ABCD  points  G,  H^ 
E,  F,  P,  etc.,  are  taken,  so  that  GA  =  AH  =  EB 
=  BF=PC,  etc.  On  the  diagonals  AC  and  BD 
points  K,  L,  M,  N,  are  laid  off  so  that  OK  =  OL 
=  0M=  ON. 

(a)  Prove  that  GAHN  =  EBFK,  etc. 

(b)  Fvovethsit  HEKON=FPLOK.  _ 
Suggestion. 

other. 


Superpose  one  figure  on  the 


24.  The  bisectors  of  the  angles  of  a  rhomboid 
form  a  rectangle ;  those  of  a  rectangle  form  a 
square. 


^ 

^ 

0 

Parquet  Flooring. 


25.   In  the  figure  ABCDEF  is  a  regular  hex-  j 
agon.     ABHG,  BCLK,  etc.,  are  squares. 

(a)  What  kind  of  triangles  are  BHK,  CLM,  ' 
etc.  ?     Prove. 

(b)  Is   the   dodecagon  (twelve-sided  polygon) 
HKLMN.  .  .  regular?     Prove. 

♦    (c)    How    many    axes    of    symmetry    has    this 

dodecagon  ? 

,^    (d)    Has  it  a  center  of  symmetry  ? 

(e)    Are  the  points  S,  F,  B,  K  collinear?     (That 
is,  do  they  lie  in  the  same  straight  line?) 


Q. 


xx^' 


■IJsi 


Tile  Pattern. 


26.  If  from  any  two  points  P  and  Q  in  the  base  of  an 
isosceles  triangle  parallels  to  the  other  sides  are  drawn, 
two  parallelograms  are  formed  whose  perimeters  are 
equal. 

27.  The  middle  point  of  the  hypotenuse  of  a  right  tri- 
angle is  equidistant  from  the  three  vertices.  (This  is  a  very 
important  theorem.) 

28.  State  and  prove  the  converse  of  the  theorem  in  Ex.  27. 


RECTILINEAR  FIGURES. 


83 


m 


29.  The  bisectors  of  the  four  interior  angles 
formed  by  a  transversal  cutting  two  parallel  lines 
form  a  rectangle. 


•  30.  The  sum  of  the  perpendiculars  from  a  point  in 
the  base  of  an  isosceles  triangle  to  the  sides  is  equal  to 
the  altitude  from  the  vertex  of  either  base  angle  on  the 
side  opposite. 

31.  Find  the  locus  of  the  middle  points  of  all  seg- 
ments joining  the  center  of  a  parallelogram  to  points  on 
the  sides.    (See  §  159,  Ex.  8.) 

32.  In  the  parallelogram  A  BCD  points  E  and  F  are 

the  middle  points  oi  AB  and  CD  respectively.     Show 

that  AF  and  CE  divide  BD  into  three  equal  segments. 

y 

33.  The  sum  of  the  perpendiculars  to  the  sides  of  an 
equilateral  triangle  from  a  point  P  within  is  the  same 
for  all  such  points  P  (i.e.,  the  sum  is  a  constant). 

Suggestion.  Prove  the  sum  equal  to  an  altitude  of 
the  triangle. 

34.  In  any  triangle  the  sum  of  two  sides  is  greater 
than  twice  the  median  on  the  third  side. 


'  35.  A  BCD  is  a  square,  and  EFGH  a  rec- 
tangle. Does  it  follow  th'dt  AAEH  ^AFCG  and 
AEBF^AHDG'^    Prove.  H 


A      E  B 

■  36.   If  a  median  of  a  triangle  is  equal  to  half  the  base,  the  vertex 
angle  is  a  right  angle. 

37.   State  and  prove  the  converse  of  the  theorem  in  Ex.  36. 


CHAPTER   II. 


STRAIGHT  LINES  AND  CIRCLES. 


Outside  Point 


180.  A  circle  (§  12)  divides  the  plane  into 
two  parts  such  that  any  point  which  does  not 
lie  on  the  circle  lies  within  it  or  outside  it. 

181.  A  line-segment  joining  any  two  points 
on  a  circle  is  called   a  chord.     A   chord 
which    passes    through    the    center    is    a 
diameter. 

182.  If  a  chord  is  extended  in  one  or 
both  directions,  it  cuts  the  circle  and  is 
called  a  secant. 

183.  A  tangent  is  a  straight  line  which  touches  a  circle 
in  one  point  but  does  not  cut  it.     An 
indefinite  straight  line  through  a  point 
outside  a  circle  is  a  secant,  a  tangent, 
or  does  not  meet  the  circle. 

184.  The  portion  of  a  circle  included 
between  any  two  of  its  points  is  called 
an  arc  (§12).  An  arc  AB  is  denoted 
by  the  symbol  AB. 

A  circle  is  divided  into  two  arcs  by  any 
two  of  its  points.  If  these  arcs  are  equal, 
each  is  a  semicircle.  Otherwise  one  is  called 
the  major  arc  and  the  other  the  minor  arc. 

Unless  otherwise  indicated  AB  means  the  minor 
arc.     In  case  of  ambiguity  a  third  letter  may  be  used,  as  arc  AmB. 

.  84 


STRAIGHT  LINES  AND   CIRCLES. 


85 


An  arc  is  said  to  be  subtended  by  the  ^Aed  Ato 

chord  which  joins  its  end-points.  Evi- 
dently every  chord  of  a  circle  subtends  two 
arcs.  Unless  otherwise  indicated  the  arc 
subtended  by  a  chord  means  the  minor  arc. 

185.  An  angle  formed  by  two  radii  is 
called  a  central  angle.  An  angle  formed 
by  two  chords  drawn  from  the  same  point 
on  the  circle  is  called  an  inscribed  angle. 

If  the  sides  of  an  angle  meet  a  circle  the 
arc  or  arcs  which  lie  within  the  angle  are 
called  intercepted  arcs. 

If  the  vertex  of  the  angle  is  within  or  on 
the  circle  there  is  only  one  intercepted  arc  ; 
if  it  is  outside  the  circle  there  are  two  inter- 
cepted arcs,  as  AB  and  CD  in  the  figure. 

186.  If  a  circle  is  partly  inside  and 
partly  outside  another  circle,  then  they  cut 
each  other. 

If  two  circles  meet  in  one  and  only  one 
point,  they  are  said  to  be  tangent. 

Arcs  of  two  circles  are  tangent  to  each  other  if 
the  complete  circles  of  which  they  form  a  part  are 
tangent  to  each  other. 

187.  Two  circles  which  can  be  made  to  coincide  are  said 
to  be  equal. 

The  word  congruent  is  unnecessary  here,  since  all  circles  are  similar, 

188.  EXERCISES. 

1.  Does  the  word  circle  as  used  in  this  book  (§  12)  mean  a  curved 
line  or  the  part  of  the  plane  inclosed  by  that  line? 

2.  In  how  many  points  can  a  straight  line  cut  a  circle  ? 

3.  In  how  many  points  can  two  circles  cut  each  other  ? 


OS 


86  PLANE  GEOMETRY. 

PRELIMINARY  THEOREMS  ON  THE  CIRCLE. 

189.  Radii  or  diameters  of  the  same  circle  or  of  equal 
circles  are  equal. 

190.  If  the  radii  or  diameters  of  two  circles  are  equal, 
the  circles  are  equal. 

191.  A  diameter  of  a  circle  is  double  the  radius. 

192.  A  point  lies  within,  outside,  or  on  a  circle,  accord- 
ing as  its  distance  from  the  center  is  less  than,  greater 
than,  or  equal  to  the  radius. 

\      193.    If  an  unlimited  straight  line  contains  a  point 
within  a  circle,  then  it  cuts  the  circle  in  two  points. 

194.  //  two  circles  intersect  once,  they  intersect  again. 
See  figure,  §  186. 

195.  //  a  straight  line  is  tangent  to  each 
of  two  circles  at  the  same  point,  then  the 
circles  do  not  intersect,  hut  are  tangent  to 
each  other  at  this  point.     See  §  186. 

196.  //  two  arcs  of  the  same  circle  or  equal  circles  can 
he  so  placed  that  their  end-points  coincide  and  also  their 
centers,  then  the  arcs  coincide  throughout  or  else  form 
a  complete  circle. 

-V      197.    If  in  two  circles  an  arc  of  one  can  he  made  to 
coincide  with  an  arc  of  the  other,  the  circles  are  equal. 

198.  A  circle  is  conveniently  referred  to  by  indicating 
its  center  and  radius. 

Thus,  O  OA  means  the  circle  whose  center  is  0  and  radius  OA. 
When  no  ambiguity  arises,  the  letter  at  the  center  alone  may  be 
used  to  denote  the  circle.     Thus,  O  C  means  the  circle  whose  center  is  C. 


STRAIGHT  LINES  AND   CIRCLES.  87 

199.   Theorem.     In  the  same  circle  or  in  equal  circles 
equal  central  angles  intercept  equal  arcs. 


Given  the  equal  circles  C  and  C  and  Z  C=Z  C. 
To  prove   that  AB  =  Jab'. 

Proof :  Place  O  C  on  Q  &  so  that  C  falls  on  &,  and 
Z  c  coincides  with  Z  C'. 

Then  A  falls  on  A'  and  B  on  b'.  (§  189) 

Hence  AB  =  A^'.  (§  196) 

200.  Theorem.  In  the  same  circle  or  in  equal  circles 
equal  arcs  are  intercepted  by  equal  central  angles. 

Given  O  C  =  O  C,  AB  =  A^.     (See  figure,  §  199.) 

To  prove  that  Zc=Zc' . 

Proof :  Since  AB  =  A'B\  the  equal  circles  can  be  made  to 
coincide  in  such  manner  that  the  arcs  will  also  coincide. 

That  is,  A  will  fall  on  A',  B  on  b\  and  C  on  C' .  Hence 
Z  c  coincides  with  Z  c'. 

201.  EXERCISES. 

1.  If  two  equal  circles  are  so  placed  that  their  centers  coincide, 
what  can  be  said  of  the  circles? 

2.  Can  two  intersecting  circles  have  the  same  center? 

"     3.   From  a  point  on  a  circle  construct  two  equal  chords. 
,       4.    Show  that  the  bisector  of  one  of  the  angles  formed  by  the  chords 
in  Ex.  3  passes  through  the  center  of  the  circle. 


88  PLANE  GEOMETRY.  /, 

202.  Measurement  of  Angles.  If  the  ^erigon  at  the  cen- 
ter of  a  circle  be  divided  by  radii  into  360  equal  angles, 
these  radii  will  divide  the  circle  into  360  equal  arcs  accord- 
ing to  the  theorem,  §  199.  Hence,  we  speak  of  an  arc 
of  1°,  2°,  3°,  etc.,  and  similarly  for  minutes  and  seconds. 

For  this  reason  a  central  angle  is  said  to  be  measured  hy 
the  arc  which  it  intercepts^  meaning  that  a  given  central 
angle  contains  a  number  of  unit  angles  equal  to  the  number 
of  unit  arcs  in  the  intercepted  arc. 

203.  Definitions.     A  quadrant  is  an  arc  of  90°. 

A  semicircle  is  an  arc  of  180°.  A  right  angle  is,  there- 
fore, measured  by  a  quadrant  and  a  straight  angle  is  meas- 
ured by  a  semicircle. 

A  sector  is  a   figure   formed    by  two      X  X^ 

radii  and  their  intercepted  arc. 

Thus,  the  sector  BCA  is  formed  by  the  radii 
CB,CA,2indiBA. 

204.  EXERCISES. 

1.  Show  how  to  bisect  an  arc,  using  §§  48,  199. 

Divide  an  arc  into  four  equal  parts.     Into  eight  equal  parts. 

2.  Show  that  in  the  same  circle  or  in  equal  circles  two  sectors  hav- 
ing equal  angles  are  congruent. 

3.  Show  that  if  two  sectors  in  the  same  circle  or  in  equal  circles 
have  equal  arcs,  the  sectors  are  congruent. 

4.  How  many  degrees  in  the  arc  which  measures 'a  right  angle? 
a  straight  angle?  half  a  right  angle?  three  fourths  of  a  straight  an- 
gle? two  thirds  of  a  right  angle?  two  thirds  of  a  straight  angle? 

5.  Show  that  the  diameter  is  the  longest  chord  of  a  circle. 

6.  Show  that  the  two  arcs  into  which  the  extremities  of  a  diameter 
divide  a  circle  are  equal ;  that  is,  each  is  a  semicircle. 

Suggestion.     Fold  the  figure  over  on  the  diameter.  ' 

7.  Show  that  by  bisecting  the  angles  between  two  perpendicular 
diameters,  a  circle  is  divided  into  eight  equal  parts. 


STRAIGHT  LINES  AND   CIRCLES. 


89 


205.   Theorem.     A  diameter  perpendicular  to  a  chord 
bisects  the  chord  and  also  its  subtended  arc. 


Given  the  diameter  EF  ±  AB  at  D. 

To  prove  that  AD  =  DB  and  AF  —  BF. 

Proof :    Draw  the  radii  CA  and  CB. 

If  it  can  be  shown  that  A  ACD  ^  A  BCD,  then 
(1)  AD=BD  (Why?),  and  (2)  Zacd=Z.bcd  (Why?); 
(3)  i>=i>  (Why?). 

206.  Theorem.     A  line  perpendicular  to  a  radius  at 
its  extremity  is  tangent  to  the  circle. 


h 


^  Given  AB  _L  CD  at  D. 

To  prove  that  AB  is  tangent  to  the  circle ;  that  is,  does 
not  meet  it  in  any  other  point  than  D. 

Proof :     Let  E  be  any  point  of  AB  other  than  D.  f^ 

Draw  segment  CE.     Then  CE  >  CD  (Why?). 

Hence  E  is  outside  the  circle  (§  192). 

That  is,  every  point  of  AB  except  D  is  outside  the  circle, 
and  hence  AB  is  tangent  to  the  circle  (§  183). 


90  /  PLANE  GEOMETRY. 

207.  Theorem.  //  a  line  is  tangent  to  a  circle,  it  is 
perpendicular  to  the  radius  drawn  to  the  point  of  tan- 
gency. 


^  D  E     B 

Given  O  CD  with  a  line  AB  tangent  to  the  circle  at  D. 

To  prove  that  AB  ±  CD  at  B. 

Proof:  If  CB  is  iiot-L^J5,  then  some  other  line,  as  CE^ 
must  be±^B  (§  67),  thus  making  CE<  CB  (Why?). 

The  point  E  would  then  lie  within  the  circle  (§  192),  and 
the  line  AB  would  meet  the  circle  in  two  points  (§  193). 

But  this  contradicts  the  hypothesis  tliat  AB  is  tangent 
to  the  circle. 

Hence  no  other  line  than  CB  can  be  perpendicular  to  AB 
from  c,  and  as  one  such  line  exists,  it  must  be  CB. 

208.  EXERCISES. 

1.  What  type  of  proof  ^s  used  in  the  preceding  paragraph  ? 

2.  How  are  the  two  theorems  immediately  preceding  related  to 
each  other. 

3.  Show  that  there  is  only  one  tangent  to  a  circle  at  a  given  point 
on  it,  and  that  the  perpendicular  from  the  center  upon  the  tangent 
meets  it  at  the  point  of  contact. 

4.  A  perpendicular  to  a  tangent  at  the  point  of  tangency  passes 
through  the  center  of  the  circle. 

5.  The  perpendicular  bisector  of  any  chord  passes  through  the 
center  of  the  circle. 

6.  Two  tangents  at  the  extremities  of  a  diameter  are  parallel. 

7.  A  diameter  bisects  all  chords  parallel  to  the  tangei^ts  at  its 
extremities,  and  also  bisects  the  central  angle  subtended  by  each  chord. 


STRAIGHT  LINES  AND   CIRCLES.    '  91 

8.  A  diameter  bisecting  a  chord  (or  its  subtended  arc)  is  perpendicular 
to  the  chord. 

9.  The  mid-points  of  parallel  chords  all  lie  on  a  diameter. 

10.    A  tangent  to  a  circle  at  the  mid-point  of  any  arc  is  parallel  to 
the  chord  of  the  arc. 

209.   Theorem.     If  two  circles  meet  on  the  line  joining 
their  centers,  they  are  tangent  to  each  other  at  this  point. 


Given  (D  CB  and  CB  meeting  in  a  point  B  on  the  line  C'C. 
To  prove  that  (D  CB  and  c' B  are  tangent  to  each  other 
at  B. 

Proof:     (1)    When  each  circle  is  outside  the  other. 

Let  D  be  ani/  point  on  O  c'b  other  than  B. 

Draw  CD  and  c'd. 

Then  CD  +  c'd  >  CB  +  c'b.  (Why  ?) 

But  c'b  =  c'd. 

.-.  CD  >  CB.  (Ax.  IX,  §  119) 

.'.  D  is  outside  of  O  CB. 

(2)    WTien  O  CB  is  inside  of  O  c'b. 

Let  D  be  ani/  point  on  O  CB  other  than  B. 

Draw  CD  and  c'd. 

Then  c'c  +  CD  >  c'd.  (Why  ?) 

But  C'C  +  CD  =  C'C  +  CB  =  c'b.  (Why  ?) 

.-.  C'b  >  c'd.  (Ax.VII,  §  82) 

.-.  D  is  within  O  c'b. 

Therefore  O  c'b  and  O  CB  have  only  one  point  in  com- 
mon and  hence  are  tangent  to  each  other  (§  186). 


92 


PLANE  GEOMETRY. 


PROBLEMS  AND  APPLICATIONS. 

1.  If  the  distance  between  the  centers  of  two  circles  is  equal  to  the 
sum  of  their  radii,  how  are  the  circles  related?     Construct  and  prove. 

2.  If  the  distance  between  the  centers  of  two  circles  is  equal  to  the 
difference  of  their  radii,  how  are  the  circles  related  ?  Construct  and 
prove  ? 

3.  If  the  distance  from  the  center  of  a  circle  to  a  straight  line  is 
equal  to  the  radius,  how  is  the  line  related  to  the  circle  ?  Construct 
and  prove. 

4.  Given  two  circles  having  the  same  center,  construct  a  circle 
tangent  to  each  of  them.  Can  more  than  one  such  circle  be  con- 
structed ?     What  is  the  locus  of  the  centers  of  all  such  circles  ? 

5.  Prove  the  converse  of  the  theorem  in  §  209. 

6.  The  straight  line  joining  the  centers  of  two  intersecting  circles 
bisects  their  common  chord  at  right  angles. 

7.  A  line  tangent  to  each  of  two  equal 
circles  is  either  parallel  to  the  segment  join- 
ing their  centers  or  else  it  bisects  this  seg- 
ment. • 

8.  In  the  figure  AD  =  DB.  Semicircles 
are  constructed  on  AD,  DB,  and  yl5  as  di- 
ameters. Which  semicircles  are  tangent  to 
each  other? 


9.  In  the  figure  A,  B,  C,  D  are  the  vertices  of  a 
square.  Show  how  to  construct  the  entire  figure. 
What  semicircles  are  tangent  to  each  other? 


This  construction  occurs  frequently  in 
designs  for  tile  flooring.  See  accompanying 
figure.     This  is  from  a  Roman  mosaic. 


STBAIGHT  LINES  AND   CIRCLES. 


93 


%       10.   Given  two  parallel  lines  BE  and  AD,  to  con- 
/  struct  arcs  which  shall  be  tangent  to  each  other  and 
one  of  which  shall  be  tangent  to  BE  at  B  and  the 
other  tangent  to  AD  at  A. 

Solution.  Draw  AB  and  bisect  this  segment  at 
C;  construct  J- bisectors  oi  AC  and  BC.  From  A 
and  B  draw  Js  to  AD  and  BE  respectively,  thus 
locating  the  points  0  and  0'. 

Prove  that  0  and  0'  are  the  centers  of  the  required 
arcs. 

Suggestion.  Show  that  0,  C,  and  0'  lie  in  a 
straight  line  and  use  the  theorem  of  §  209. 

This  construction  occurs  in  archi- 
tectural designs  and  in  many  other 
applications.  In  the  accompanying 
designs  pick  out  all  the  arcs  that  ^^ 
are  tangent  to  each  other  and  also 
the  points  of  tangency.  ^  S^r^„  ^^^^ 

11.  On  the  sides  of  the  equilateral  triangle  ABC  as  diameters, 
semicircles  are  drawn,  as  AEFB.  Also  with  A,  B,  C  as  centers  and 
AB  B,s  radius  arcs  are  drawn,  ^ 

as^^,  ^. 

(a)  Prove  that  the  arcs 
AEFB,  BDEC,  and  CFDA 
meet  in  pairs  at  the  middle 
points  D,  E,  F  of  the  sides  of 
the  triangle. 

Suggestion.  If  the  mid- 
dle points  of  the  sides  of  an 
equilateral  triangle  are  joined, 
what  kind  of  triangles  are  formed? 

(h)    What  arcs  in  this  figure  are  tangent  to  each  other  ? 
•  (c)    Has  the  figure  one  or  more  axes  of  symmetry? 

This  figure  and  the  two  follo^ipg  occur  frequently  in 
church  windows  and  other  decorative  designs. 


Fourth  Presbyterian  Church,  Chicago. 


94 


PLANE  GEOMETRY. 


11.  Construct  the  design  shown  in  the  figure. 

Suggestion.  Divide  the  diameter  AB  into  six  equal  parts  and 
construct  the  three  semicircles. 

On  DC  and  DC  as  bases  construct  equilateral 
triangles  with  vertices  0  and  0'. 

With  radius  equal  to  CB  and  centers  O  and 
0'  construct  circles.  AC'        D  C    B 

(a)  Prove  that  O  0  is  tangent  to  each  of  the  three  semicircles. 
Likewise  O  0'. 

(h)   Erect  a,  ±  to  AB  at  D  and  prove  ©  0  and  0'  tangent  to  it. 

(c)    Prove  circles  with  centers  at  0  and  0'  tangent  to  each  other. 
0  (c?)    Has  this  figure  one  or  more  axes  of  symmetry? 

12.  In  the  figure  AB,  CD  and  OD  are  bisected,  and  O'O"  II  AB 
through  E.    DO'  =  DO"  =  |  DB.    Circles  are 
constructed  as  shown  in  the  figure. 

(a)  If  ^5  is  4  feet,  what  is  the  radius  of 
each  circle?  LQ"\     E\ 

(b)  Prove  that  O  0  is  tangent  to  GO'  and 
also  to  O  0". 

Suggestion.  Show  that  00'  is  the  sum  of  the  radii  of  the  two 
circles. 

(c)  Is  Q  0'  tangent  to  the  arc  ACB  and  also  to  the  line  AB  ? 

(d)  Has  this  figure  one  or  more  axes  of  symmetry  ? 

13.  A  BCD  is  a  square.  Arcs  are  constructed  with  A,  B,  C,  D  as 
centers  and  with  radii  each  equal  to  one  half  the  side  of  the  square- 
The  lines   AC,  BD,   MN,   and  RS  are 

drawn,  and  the  points  E,  F,   G,  H  are 
connected  as  shown  in  the  figure. 

The  arc  SN  is  extended  to  P,  forming 
a  semicircle.  The  line  LP  meets  SAf  in 
K,  and  BK  meets  MN  in  0'. 

(a)    Prove  that  EFGH  is  a  square. 

{h)  Prove  that  A  KLO'  and  KPB  are 
mutually  equiangular  and  each  isosceles. 

(c)    Prove  that  0  O'K  is  tangent  to  FG  and  to  SN. 
^  (c?)    How  many  axes  of  symmetry  has  the  solid  figure? 

(e)  Show  that  0  O'K  is  tangent  to  RN  by  drawing  O'C  and  fold- 
ing the  figure  over  on  the  axis  of  symmetry  MN. 


\  \ 


STRAIGHT  LINES  AND   CIRCLES. 


95 


210.  Theorem.     An  angle  inscribed   in  a  circle  is 
measured  by  one  half  the  intercepted  arc. 


Given  Z  DBA  inscribed  in  O  CB. 
To  prove  that  Z  DBA  is  measured  by  ^  AD. 
Proof  :    (1)    If  one  side^  as  BD^  is  a  diameter. 
Draw  the  radius  CA.     Show  that  Z  2  =  |  Z  1. 
But  Z  1  is  measured  by  AD  (§  202). 
Hence  Z  2  is  measured  by  ^  AD. 

(2)  If  the  center  C  lies  within  the  angle. 
Draw  the  diameter  BE. 

Now  Z  DBA  =  Z  1  +  Z  2. 
Complete  the  proof. 

(3)  If  the  center  C  lies  outside  the  angle. 

Draw  BE  and  use  the  equation  Z  DBA  =  Z\  -  Z2. 

211.  It  follows  from  §  210  that  if  in  equal  circles  two 
inscribed  angles  intercept  equal  arcs,  they  are  equal ;  and 
conversely,  that  if  equal  angles  are  inscribed  in  equal 
circles,  they  intercept  equal  arcs. 


212. 


EXERCISES. 


1.  If  the  sides  of  two  angles  BAD  and  BA'D  pass  through  the 
points  B  and  Z)  on  a  circle,  and  if  the  vertex  A  is  on  the  minor  arc 
BD  and  A'  is  on  the  major  arc  BD,  find  the  sum  of  the  two  angles. 

2.  In  Ex.  1  if  the  points  B  and  D  remain  fixed  while  the  vertex  A 
of  the  angle  is  made  to  move  along  the  minor  arc  of  the  circle,  what 
can  be  said  of  the  angle  A  ?    What  if  it  moves  along  the  major  arc  ? 


96  PLANE  GEOMETRY. 

213.  Theorem.  The  locus  of  the  vertices  of  all  right 
triangles  on  a  given  hypotenuse  is  a  circle  whose  diame- 
ter is  the  given  hypotenuse.  ^  ^ ^' 

Outline  of  Proof :     Let  AB  be  the 

given  hypotenuse. 

(1)  If  P  is  any  point  on  the  circle 
whose  diameter  is  AB^  Z  APB  =  rt.  Z. 
(Why  ?) 

'(2)  .If  AF^ B  is  any  right  triangle  with  AB  as  hypote- 
nuse, then  AC=CB  =  CP'.     (See  Ex.  27,  p.  82.) 

State  the  proof  in  full. 

214.  PROBLEMS  ON  LOCI. 

Find  the  following  loci : 

1.  The  centers  of  all  circles  of  fixed  radius  tangent  to  a  fixed  line. 

2.  The  centers  of  all  circles  tangent  to  two  parallel  lines. 

3.  The  centers  of  all  circles  tangent  to  both  sides  of  an  angle. 

4.  The  centers  of  all  circles  tangent  to  a  given  line  at  a  given 
point.     Is  the  given  point  a  part  of  this  locus  ? 

5.  The  vertices  of  all  triangles  which  have  a 
common  base  and  equal  altitudes. 
I    6.    The  middle  points  of  all  chords  through  a 

fixed  point  on  a  circle.     Use  Ex. 
8,  §208,  and  then  §213. 

7.  The  points  of  intersection 
of   the    diagonals    of    trapezoids 
formed  by  the  sides  of  an  isosceles  triangle  and  lines 
parallel  to  its  base. 

8.  Two  vertices  of  a  triangle  slide  along  two 
parallel  lines.  What  is  the  locus  of  the  third  vertex  if  the  triangle  is 
fixed  in  size  and  shape? 

,.     *,  9.   ^5CZ)  is  a  parallelogram  all  of  whose  sides  are  of  fixed  length. 
*The  side  AB  is  fixed  in  position.     Find  the  locus  of  the  middle  points 
of  the  remaining  three  sides. 


STRAIGHT  LINES  AND   CIRCLES. 


97 


10.  Prove  that  in  the  same  circle  or  in  equal  circles  equal  chords  are 
equally  distant  from  the  center. 

Suggestion.  MB  =  ND.  Why?  Then  prove 
A  BMC  =  A  CND. 

11.  State  and  prove  the  converse  of  the  theorem 
in  the  preceding  exercise.  (What  parts  of  A  BMC 
and  CND  are  now  known?) 

12.  Find  the  locus  of  the  middle  points  of  all 
chords  of  equal  length  in  the  same  circle. 

13.  Find  the  locus  of  the  middle  point  of  a  segment  AB  of  fixed 
length  which  moves  so  that  its  end-points  slide  along  the  sides  of  a 
right  angle.     (Use  Ex.  27,  p.  82.) 

14.  Find  the  locus  of  the  points  of  contact  of  two  varying  circles 
tangent  to  each  other,  and  each  tangent  to  a 
given  line  at  a  given  point. 

Suggestion.     A  and  B  are  tlie  fixed  points,     I         \\z:'-i^.^S' 
and  P  one  point  of  contact  of  the  circles.     Draw 
the   common  tangent   PD.      Prove  AD  =  DP 
and  DB  =  DP. 

Hence,  D  is  the  middle  point  of  AB  and  DP  is  constant.  That  is, 
the  locus  is  a  circle  of  which  AB  is  a  diameter. 

15.  Find  the  locus  of  the  centers  of  all  circles  tangent  to  a  fixed 
circle  at  a  fixed  point  P.  Is  the  fixed  point  P  a  part  of  this  locus  ? 
Is  the  center  of  the  fixed  circle  a  part  of  it? 

^     16.    Find  the  locus  of  the  centers  of  all  circles  of  the  same  radius 
which  are  tangent  to  a  fixed  circle. 
%       Under  what  conditions  will  this  locus  include  the  fixed  circle  itself? 
The  center  of  this  fixed  circle  ? 

Will  the  locus  ever  contain  a  circle  within  the  fixed  circle? 
'^     Under  what  conditions  will  the  locus  consist 
,    of  two  circles,  each  outside  the  fixed  circle  ? 
i       Under  what  condition  does  the  locus  consist 
V,pf  only  one  circle? 

17.  In  making  core-boxes,  pattern  makers 
use  a  square  as  indicated  in  the  figure  to  test 
whether  or  not  the  core  is  a  true  semicircle.  Is 
this  method  correct?     Prove. 


98  PLANE  GEOMETRY. 

215.  Theorem.  The  arcs  intercepted  by  two  parallel 
chords  or  by  a  tangent  and  a  chord  parallel  to  it  are 
equal. 


Given  AB  \\  DE  and  LK  II  MN. 

To  prove  that  AD  —  BE  and  LB  =  KB. 

Proof :     (1)  Draw  chord  DB. 

Compare  Z  1  and  Z  2,  and  hence  show- 
that  iB  =  B^.  (Why?) 

(2)  Draw  the    radius  CB   to   the   point   of   tangency. 
Then  CB  ±  MN  and  CB  J.  LK.     (Why?) 

Prove  A  LCS  ^  A  KCS,  and  hence  that  LB  =  BK.  (§  199) 

216.  EXERCISES. 

1.  Prove  that  a  tangent  at  the  vertex  of  an  inscribed  angle  forms 
equal  angles  with  the  two  sides,  if  these  are  equal  chords. 

2.  If  the  vertices  of  a  quadrilateral  lie  on  a  circle,  any  two  of  its 
opposite  angles  are  supplementary. 

3.  If  two  chords  of  a  circle  are  perpendicular  to  each  other,  find 
the  sum  of  each  pair  of  opposite  arcs  into  which  they  divide  the  circle. 

4.  If  the  vertices  of  a  trapezoid  lie  on  a  circle, 
its  diagonals  are  equal. 

5.  Two  circles  intersect  at  C  and  D.  Diame- 
ters CA  and  CB  are  drawn.  Prove  that  A,  D,  B 
lie  on  a  straight  line. 

Suggestion.     Prove  that  ZADC  =  Z  CDB  =  rt.  Z. 


STRAIGHT  LINES  AND   CIRCLES. 


99 


217.   Theokem.     An  angle  formed  by  two  intersecting   ^ 
chords  is  measured  by  one  half  the  sum  of  the  arcs  inter- 
cepted by  the  angle  itself  and  its  vertical  angle. 


Given  Z 1  formed  by  the  chords  AB  and  DE. 

To  prove  that  Zl  is  measured  by  |  (^AE+  BD). 

Proof  :     Through  A  draw  the  chord  AF  ||  ED. 

Compare  Z  1  and  Z  3. 

Compare  AE  and  DF,  also  AE  +  BD  and  BD  +  DF. 

How  is  Z  3  measured  ? 

Hence,  how  is  Z 1  measured  ? 


218. 


EXERCISES. 


1.  A  chord  AB  ia  divided  into 
three  equal  parts,  AC,  CD,  and 
DB.  OA,  OC,  OD,  and  OB  are 
drawn.  Compare  the  angles  AOC, 
COD,  and  DOB. 

2.  The  accompanying  table 
refers  to  the  figure  in  §  217.  Fill 
out  blank  spaces. 

3.  In  a  circle  C  with  a  diameter 
AB  21,  chord  AD  \q  drawn,  and  a 
radius  CE  \\  A  D.  Prove  that  arcs 
DE  and  EB  are  equal. 

4.  The  vertices  of  a  square  A  BCD  all  lie  on  a  circle.  E  is  any 
point  on  the  arc  AB.  Prove  that  EC  and  ED  divide  the  angle  A  EB 
into  three  equal  parts. 


Zl 

AE 

BD 

EB 

AnD 

35° 

40° 

80° 

48° 

50 

216 

40° 

50° 

60° 

60° 

54° 

190° 

45° 

90° 

180° 

34° 

108° 

164° 

100 


PLANE  GEOMETRY, 


219.  Theorem.  An  angle  formed  by  a  tangent  and 
a  chord  drawn  from  the  'point  of  tangency  is  measured 
by  one  half  the  intercepted  arc. 

D/ 


Given  Z 1  formed  by  tangent  BD  and  chord  BA. 

To  prove  that  Z  1  is  measured  hy  ^  BA. 

Proof :     Draw  a  chord  EF 11  BD  intersecting  BA  in  Q. 

Compare  Z  1  and  Z  2,  also  EB  and  BF. 

How  is  Z  2  measured  ? 

Hence,  how  is  Z 1  measured  ? 

Give  the  proof  in  full. 

220.  Definitions.  A  segment  of  a  circle,  or  a  circle-segf- 
ment,  is  a  figure  formed  by  a  chord  and  the  arc  which  it 
subtends.  For  each  chord  there  are  two 
circle-segments  corresponding  to  the  two 
arcs  which  it  subtends. 

If  a  chord  is  a  diameter  the  two  circle- 
segments  are  equal. 

An  angle  is  said  to  be  inscribed  in  an  arc  if  its  vertex 
lies  on  the  arc  and  its  sides  meet  the  arc  in 
its  end-points. 

Such  an  angle  is  also  said  to  be  inscribed     a^ 
in  the  circle-segment  formed  by  the  arc  and 
its  chord. 

E.g.  Z 1  is  inscribed  in  the'  arc  APB  or  in  the  segment  APB. 


STRAIGHT  LINES  AND   CIRCLES. 


101 


221.  EXERCISES. 

1.  Show  that  an  angle  inscribed  in  a  semicircle  is-j;  right  angle. 

2.  If  the  sides  of  a  right  angle  pass  through  the  extremities  of  a 
diameter,  show  that  its  vertex  lies  on  the  circle. 

3.  If  a  triangular  ruler  J/iVO,  right-angled  at  0,  is  moved  about  in 
the  plane  so  that  two  fixed  points,  A  and  B,  lie  always 

on  the  sides  MO  and  NO  respectively,  what  path  o 

does  the  point  0  trace? 

4.  Draw  two  concentric  circles,  having  different 
radii,  and  show  that  all  chords  of  the  outer  circle 
which  are  tangent  to  the  inner  circle  are  equal. 

5.  In  an  equilateral  triangle  construct  three  equal  circles,  each 
tangent  to  the  two  other  circles  and  to  two  sides 
of  the  triangle. 

Suggestion.  Construct  the  altitudes  of  the 
triangle  and  bisect  angles  as  shown  in  the  figure. 
Complete  the  construction  and  prove  that  the 
figure  has  the  required  properties. 

(a)   Has  the  figure  consisting  of  the  triangle 
and  the  three  circles  one  or  more  axes  of  symmetry? 
u    (b)  Has  it  a  center  of  symmetry? 

6.  Within  a  given  circle  construct  three  equal  circles,  each  tangent 
to  the  other  two  and  to  the  given  circle. 

Suggestion.     Trisect  the  circle  at  D,  E,  c 

and  F  by  making  angles  at  the  center  each 
equal  to  120°.  Draw  tangents  at  D,  E,  and 
F,  and  prove  that  A  ABC  is  equilateral. 

Construct  the  altitudes  and  prove  that  they 
meet  the  sides  of  the  triangle  at  the  points  of 
tangency  of  the  given  circle  with  the  sides  of 
ABC,  and  also  that  they  pass"  through  the 
center  of  the  given  circle. 

Bisect  angles  as  shown  in  the  figure  and  prove  that  the  centers  of 
the  required  circles  are  thus  obtained. 

•     (a)    Has  the  figure  consisting  of  the  four  circles  one  or  more  axes 
of  symmetry? 
».  (6)    Has  it  a  center  of  symmetry  ? 


102 


PLANE  GEOMETBT. 


\ 

22ii.  Theorem.  The  angle  formed  hy  two  secants, 
two  ia;ng0nts;  x)r:a  tangent  and  a  secant,  meeting  outside 
a  circle,  is  measured  hy  one  half  the  difference  of  the 
intercepted  arcs. 


Outline  of  Proof :  In  each  case  the  given  angle  is  equal 
to  Z  1,  and  the  arc  which  measures  Z  1  is  the  difference 
between  two  arcs,  one  of  which  is  the  larger  of  the  two 
intercepted  arcs  and  the  other  is  equal  to  the  smaller. 
For  instance,  in  the  first  figure, 

FH  =  DF  ~  DH  =z  DF  —  BE. 
,     Give  the  proof  in  detail  for  each  figure. 

223.  EXERCISES. 

1.  If  (in  left  figure,  §222)  ZA  =17°  and  EB  =  25 ^  find  DF. 

2.  If  ZA  =37°  (in  middle  figure),  find  the  arcs  into  which  the 
points  B  and  E  divide  the  circle. 

3.  With  a  given  radius  construct  a  circle  passing  through  a  given 
point.  How  many  such  circles  can  be  drawn  ?  What  is  the  locus  of 
the  centers  of  all  such  circles  ? 

4.  Draw  a  circle  passing  through  two  given  fixed  points.  How 
many  such  circles  are  there  ?  What  is  the  locus  of  the  centers  of  all 
such  circles  ? 

5.  Construct  a  circle  having  a  given  radius  and  passing  through 
two  given  points.  How  many  such  circles  can  be  drawn?  Is  this 
construction  ever  impossible?  Under  what  conditions  is  only  one 
such  circle  possible  ? 


STRAIGHT  LINES  AND   CIRCLES.  103 

224.   Problem.     To  construct  a  circle   through   three 
fixed  points  not  all  in  the  same  straight  line. 


Given  three  points  A,  B,  C  not  in  the  same  straight  line. 
To  construct  a  circle  passing  through  them. 
Construction.     Let  the  student  give  the  construction  and 
proof  in  full.     (See  §  132.) 

225.  Definition.  The  circle  OA  in  §  224  is  said  to  be 
circumscribed  about  the  triangle  ABC  and  the  triangle  is 
said  to  be  inscribed  in  the  circle. 

226.  EXERCISES. 

1.  In  the  construction  of  §  224  why  do  DM  and  EN  meet  ? 

2.  Why  cannot  a  circle  be  drawn  through  three  points  all  lying  in 
the  same  straight  line  ?    Make  a  figure  to  illustrate  this. 

3.  Show  that  an  angle  inscribed  in  an  arc  is  greater  than  or  less 
than  a  right  angle  according  as  the  arc  in  which  it  is  inscribed  is  less 
than  or  greater  than  a  semicircle. 

4.  Prove  that  the  bisectors  of  the  angles  of  an  equilateral  triangle 
pass  through  the  center  of  the  circumscribed  circle. 

5.  Draw  a  circle  tangent  to  two  fixed  lines.  How  many  such 
circles  are  there  ?  What  is  the  locus  of  their  centers  ?  Is  the  point 
of  intersection  part  of  this  locus?     Discuss  fully. 

6.  Show  that  not  more  than  one  circle  can  be  drawn  through  three 
given  points,  and  hence  that  two  circles  which  coincide  in  three 
points  coincide  throughout. 


104  PLANE  GEOMETRY. 

227.  Problem.  To  construct  a  circle  tangent  to  each 
of  three  lines,  no  two  of  which  are  "parallel  and  not  all 
of  which  pass  through  the  same  point. 

Given  the  lines  /i,  h,  ly 

To  construct  a  circle  tangent  to  each  of  these  lines. 

Construction.  Since  no  two  of  the  lines  are  II,  let  \  and 
Zg  meet  in  ^,  l^  and  Zg  in  ^,  and  Zg  and  Zj  in  D,  where  A^  />♦, 
and  Z)  are  distinct  points. 

Draw  the  bisectors  of  Z^  and  AB  and  let  them  meet 
in  point  C. 

Then  C  is  the  center  of  the  required  circle.  (See  §  131.) 
Give  the  proof  in  full. 

228.  Definitions.  The  circle  in  the  construction  of  §  227 
is  said  to  be  inscribed  in  the  triangle  ABB. 

Three  or  more  lines  which  all  pass  through  the  same  point 
are  called  concurrent.  Hence  the  lines  Z^,  Zg,  Zg  are  not  con- 
current. 

229.  EXERCISES. 

1.  Why  is  the  construction  of  §  227  impossible  if  /^  Z2,  and  h  are 
concurrent  ? 

2.  If  two  of  the  lines  are  parallel  to  each  other,  show  that  the  con- 
struction is  possible.  How  many  tangent  circles  can  be  constructed 
in  this  case  ?  Draw  a  figure  and  give  the  construction  and  proof  in 
full. 

3.  Is  the  construction  possible  when  all  three  lines  are  parallel? 
Why? 


STRAIGHT  LINES  AND   CIRCLES.  105 

4.  If  two  sides  of  the  triangle  are  produced,  as  AB  and  AD  in  the 
figure  of  §  227,  construct  a  circle  tangent  to  the  side  BD  and  to  the 
prolongations  of  the  sides  AB  and  AD. 

This  is  called  an  escribed  circle  of  the  triangle. 

5.  How  many  circles  can  be  constructed  tangent  to  each  of  three 
straight  lines  if  they  are  not  concurrent  and  no  two  of  them  are 
parallel? 

6.  Draw  a  triangle  and  construct  its  inscribed  and  circumscribed 
circles  and  its  three  escribed  circles. 

230.  Pkoblem.  From  a  given  point  outside  a  circle 
to  draw  a  tangent  to  the  circle. 


n 

Given  O  CA  and  an  outside  point  P. 

To  construct  a  tangent  from  P  to  the  circle. 

Construction.  Draw  CP.  On  CP  as  a  diameter  con- 
struct a  circle,  cutting  the  given  circle  in  the  points 
A  and  B. 

Draw  tlie  lines  PA  and  PB. 

Then  PA  and  PB  are  both  tangents. 

Give  the  proof. 

231.  EXERCISES. 

1.  If  in  the  figure  of  §  2.30  the  point  P  is  made  to  move  towards 
the  circle  along  the  line  PC  until  it  finally  reaches  thgjcircle,  while 
PA  and  PB  remain  tangent  to  the  circle,  describe  the  motion  of  the 
points  A  and  B  and  also  of  the  lines  PA  and  PB.  How  does  this 
agree  with  the  fact  that  through  a  point  on  the  circle  there  is  only 
one  tangent  to  the  circle  ? 


106  PLANE  GEOMETRY. 

2.  Can  a  tangent  be  drawn  to  a  circle  from  a  point  inside  the 
circle  ?     Why  ? 

3.  Show  that  the  line  connecting  a  point  outside  a  circle  with  the 
center  bisects  the  angle  formed  by  the  tangents  from  that  point. 

4.  Why  are  not  more  than  two  tangents  possible  from  a  given 
point  to  a  circle  ? 

5.  The  two  tangents  which  can  be  drawn  to  a  circle  from  an  ex- 
terior point  are  equal. 

^     6.    In  a  right  triangle  the  hypotenuse  plus  the  diameter  of  the 
inscribed  circle  is  equal  to  the  sum  of  the  two  legs  of  the  triangle. 

7.  If  an  isosceles  triangle  inscribed  in  a  circle  has  each  of  its  base 
angles  double  the  vertex  angle,  and  if  tangents  to  the  circle  are 
drawn  through  the  vertices,  find  the  angles  of  the  resulting  triangle. 

8.  If  the  angles  of  a  triangle  ABC  inscribed  in  a  circle  are  64°, 
72°,  and  44°,  find  the  angles  of  the  triangle  formed  by  the  tangents  to 
the  circle  at  the  points  A ,  B,  and  C 

SUMMARY  OF   CHAPTER  II. 

1.  Make  a  list  of  all  the  definitions  involving  the  circle. 

2.  State  the  theorems  on  the  measurement  of  angles  by  inter- 
cepted arcs. 

3.  State  the  theorems  involving  equality  of  chords,  central 
angles,  and  intercepted  arcs. 

4.  State  the  theorems  on  the  tangency  of  straight  lines  and 
circles. 

5.  State  the  theorems  involving  the  tangency  of  two  circles. 

6.  Make  a  list,  to  supplement  that  in  the  summary  of  Chapter  T, 
of  ways  in  which  two  angles  or  two  line-segments  may  be  proved  equal. 

7.  State  the  ways  in  which  two  arcs  of  the  same  or  equal  circles 
may  be  proved  equal. 

8.  State  the  problems  of  construction  given  in  Chapter  II. 

v^/  9.    Explain  what  is  meant  by  saying  that  a  central  angle  is  meas- 
ured by  its  intercepted  arc. 

10.    State    some    of    the    important   applications  of   Chapter  11. 
(Return  to  this  question  after  studying  those  which  follow.) 


STRAIGHT  LINES  AND   CIRCLES. 


107 


PROBLEMS  AND  APPLICATIONS. 

1.  Given  two  roads  of  different  width  at  right  angles  to  each 
other,  to  connect  them  by  a  road  whose  sides  are 
arcs  of  circles  tangent  to  the  sides  of  the  roads. 

(a)  Make  the  construction  shown  in  the  figure  and 
prove  that  it  has  the  required  properties. 

ih)  Is  this  construction  possible  when  the  given 
roads  are  not  at  right  angles  to  each  other  ?    Illustrate. 

(c)  Can  the  curve  be  made  long  or  short  at  will  ? 

id)  Make  the  construction  if  the  given  roads  have  the  same  width. 

2.  Two  circles  C  and  C  are  tangent  at  the 
point  D.  ^^  is  a  segment  through  D  tern)i- 
nating  in  the  circles.  Prove  that  the  radii 
CA  and  CB  are  parallel. 

^  3.  Through  a  point  on  the  bisector  of  an 
angle  to  construct  a  circle  tangent  to  both 
sides  of  the  angle. 

Construction.  Through  the  given  point 
P  draw  EP  _L  to  ^P.  Lay  off  EB  =  EP  and 
at  D  construct  DC  ±AD  meeting  the  line  A  P 
in  C.  Then  C  is  the  center  of  the  required 
circle  and  CD  is  its  radius. 

Proof  :  Draw  PD  and  prove  that  A  DEP 
is  isosceles  and  hence  also  A  PDC. 

Is  it  possible  to  construct   another  circle  having  the   properties 
required?    If  so,  construct  it. 

This  construction  is 
used  in  the  accompany- 
ing design  in  which  the 
shape  is  determined  by 
fixing  the  pgint  P  in  ad- 
vance. 

Such    designs    are    of 
frequent  occurrence  in  decorative  work  such  as  the  steel 
ceiling  panel  given  here. 


108 


PLANE  GEOMETRY. 


4.  In  an  isosceles  triangle  construct  three  circles 
as  shown  in  the  figure. 

Suggestion.     First    construct     the    inscribed 
circle  with  center  O.     Let  the  bisector  oi  Z.A  meet    a 
this  circle  in  a  point  P.     Then  use  Ex.  3. 

5.  The  angles  formed  by  a  chord  and  a  tangent  are  equal  respec- 
tively to  the  angles  inscribed  in  the  arcs  into  which  the  end-points  of 
the  chord  divide  the  circle. 

6.  If  a  triangle  whose  angles  are  48°,  56°,  and  76°  is  circumscribed 
about  a  circle,  find  the  number  of  degrees  in  the  arcs  into  which  the 
points  of  tangency  divide  the  circle. 

7.  Divide  each  side  of  an  equilateral  triangle  into  three  equal 
parts  (Ex.  5,  §  159)  and  connect  points  as  shown  in  the  figure. 

Prove  that  DEFGHK  is  a  regular  hexagon. 

8.  If  a  circle  is  inscribed  in  the  triangle  of  Ex.  7, 
prove  that  all  sides  of  the  hexagon  are  tangent  to  the 
circle. 

Suggestion.  Show  that  the  perpendicular  bisec- 
tors of  the  segments  HK,  KD,  DE  meet  in  a  point 
equidistant  from  these  segments.  ^ 

9.  Within  a  given  square  construct  four  equal 
circles  so  that  each  circle  is  tangent  to  one  side  of 
the  square  and  to  two  of  the  circles. 

Suggestion.  First  construct  the  diagonals  of 
the  square. 

In     the 

F 


10. 


figure, 
A  BCD  is  a  rectangle  D^ 
with  AD  =  I  AB.  E 
and  F  are  the  middle 
points  oi  AB  and  CD 
respectively. 

Semicircles  are  con- 
structed with  E  and  F 
as  centers  and  \  AE  di& 
a  radius,  etc. 


Ji  E  B 

Fan  vaulting  from  Gloucester  Cathedral,  England. 


8TBAIGHT  LINES  AND   CIRCLES. 


109 


(a)  Prove  that  these  quadrant  arcs  are  tangent  to  each  other  in 
pairs  and  also  to  the  semicircles. 

(b)  Lines  are  drawn  tangent  to  the  arcs  at  the  points  where  these 
are  met  by  the  diagonals  of  the  squares  AEFD  and  BCFE.  Prove 
that  these  lines  form  squares  KLMN  and  X  YZ  W. 

(c)  Construct  the  small  circles  within  each  of  these  squares. 

The  above  design  occurs  in  fan-vaulted  ceilings. 

The  gothic  or  pointed  arch  plays  a  conspicuous  part  in 
modern  architecture,  and  examples  of  it  may  be  found  in  al- 
most any  city.    Its  most  common  use  is  in  church  windows. 

The  figure  represents  a  so-called  equilat- 
eral gothic  arch.  The  arcs  AC  and  BC  are 
drawn  from  B  and  A  as  centers  respectively, 
and  with  AB  as  a  radius. 

The  segment  ^B  is  called  the  spaw  of  the 
arch,.,  and  the  point  C  its  a'pex. 

11.  In  the  figure  AB  =  DB.  ABC,  ADE,  and  DBF  are  equi- 
lateral gothic  arches. 

(a)  Construct  the 
circle  with  center  0 
tangent  to  the  four 
arcs  as  shown. 

Suggestion.  Take 
X  so  that  DX  =  XB. 
With  centers  A  and  B 
and  radius  AX  draw 
arcs  meeting  at  0. 

Complete  the  con- 
struction and  prove  that  the  figure  has  the  required  properties. 

(b)  Prove  that  DE  and  DF  are  tangent  to  each  other.  Also  BF 
and  BC,  and  AE  and  AC. 

(c)  What  axis  of  symmetry  has  this  figure  ? 

12.  A  triangle  ABC  whose  angles  are  45°,  80°,  and  55°  is  inscribed 
in  a  circle.  Find  the  angles  of  the  triangle  formed  by  the  tangents 
at  A,  B,  and  C. 


Door,  Union  Park  Church,  Chicago. 


110 


PLANE  GEOMETRY. 


13.  Inscribe  a  circle  in  an  equilateral  gothic  arch  ABC. 

Suggestions.  Construct  CD  ±  to  AB  and  ex- 
tend it  to  P,  making  DP  =  AB.  From  P  con- 
struct a  tangent  to  A  C  at  L. 

(a)  Prove  that  A  BDP  ^  A  BLP  and  hence 
PL  =  BD. 

(b)  A  OLP  ^ABDO  and  hence  OD  =  OL. 

Then  O  OD  is  the  required  circle.     See  §  209. 

Notice  that  this  figure  is  symmetrical  with  re- 
spect to  the  line  PD,  and  hence  if  the  circle  is  proved  tangent  to  ACj 
we  know  at  once  that  it  is  tangent  to  BC. 

14.  In  the  figure  ABC  is  an  equilateral 
gothic  arch  with  a  circle  inscribed,  as  in  Ex.  1.3. 

(a)  Construct  the  two  equilateral  arches  GHE 
and  HKF,  as  shown  in  the  figure. 

Construction.  Draw  BK  and  AG  1.  AB. 
With  a  radius  equal  to  OD  -\-  DB,  and  with  0 
as  center  draw  arcs  meeting  BK  and  AG  in  K 
and  G  respectively.  Draw  GK,  construct  the 
arches  and  show  that  each  is  tangent  to  the  circle. 

(b)  Do  the  points  E  and  F  lie  on  the  circle  ? 

Suggestion.  Suppose  KF  to  be  drawn,  and 
compare  Z  HKF  with  Z  HKO  by  comparing  the 
sides  HK  and  KF  and  also  GK  and  KO. 

15.  Construct  an  arc  passing  through  a  given 
point  B,  and  tangent  to  a  given   line  AD  at 
given  point  D. 

16.  In  the  figure  ABC  is  an 
equilateral  arch.  BK  is  f  of  BD. 
KBF  and  A  HE  are  equal  equi- 
lateral arches.  Arcs  KQ  and  HQ 
are  tangent  to  arcs  KF  and  HE 
respectively. 


From  Lincoln  Cathedral,  England. 


(a)  Find  by  construction  the  center  0  of  the  circle  tangent  to  A  C, 
BC,  KF,  and  HE,  and  give  proof. 


STRAIGHT  LINES  AND   CIRCLES. 


Ill 


(b)  Find   by  construction  the  centers  of  the   arcs  KQ  and   HQ. 
How  is  this  problem  related  to  Ex.  15? 

17.  Two  circles  are  tangent  to  each  other  in- 
ternally. Find  the  locus  of  the  centers  of  all  circles 
tangent  to  both  externally. 

18.  Two  circles  are   tangent   to   each   other  ex- 
ternally.    Find  the  locus  of  the  centers  of  all  circles    f({ 
tangent  to  both,  but  external  to  one  and  internal  to 
the  other. 


19.  Two  equal  circles  are  tangent  to  each  other 
externally.  Find  the  locus  of  the  centers  of  all 
circles  tangent  to  both. 

20.  AD'B  is  an  angle  whose  vertex  is  outside 
the  circle  and  whose  sides  meet  the  circle  in  the 
points  A  and  B,  while  Z.ADB  is  an  inscribed 
angle  intercepting  the  arc  AB.  Prove  that 
Z.  ADB  >  Z.  AD'B  J  provided  each  of  the  segments 
D'A  and  D'B  cuts  the  circle  at  a  second  point. 

21.  Through  two  given  points  A  and  5  con- 
struct a  circle  tangent  to  a  given  line  which  is 
perpendicular  to  the  line  AB. 

Is  this  construction  possible  if  the  given  line 
passes  through  either  of  the  points  A  or  5?  If  it 
meets  AB  between  these  points? 

22.   In  kicking  a  goal  after  a  touchdown  in  the 
game  of  football,  the  ball  is  brought  back  into 
the  field  at  right   angles    to  the    line    marking 
the  end  of  the  field.    The  distance  between  the 
goal  posts  being  given,  and  also  the  point  at 
which  the  touchdown  is  made,  find  by  a  geo- 
metrical construction  how  far  back  into  the 
field  the  ball  must  be  brought  in  order  that 
the  goal  posts  may  subtend  the  greatest  pos- 
sible angle. 


CHAPTER  III. 

THE  MEASUREMENT   OP  STRAIGHT  LINE- 
SEGMENTS. 

232.  A  straight  line-segment  is  said  to  be  exactly  meas- 
ured when  we  find  how  many  times  it  contains  a  certain 
other  segment  which  is  taken  as  a  unit.  The  number  thus 
found  is  called  the  numerical  measure,  or  the  length  of 
the  segment. 

E.g.  a  line-segment  is  9  in.  long  if  a  segment  1  in.  long  can  be 
laid  off  on  it  9  times  in  succession. 

Thus,  9  is  the  numerical  measure,  or  the  length  of  the  segment, 
when  1  in.  is  taken  as  a  unit. 

233.  In  selecting  a  unit  of  measure  it  may  happen  that 
it  is  not  contained  an  integral  number  of  times  in  the  seg- 
ment to  be  measured. 

Thus,  in  measuring  a  line-segment  the  meter  is  often  a  convenient 
unit.  Suppose  it  has  been  applied  five  times  to  the  segment  AB  and 
that  the  last  time  the  end  falls  on  A^,  A^B  being  less  than  one  meter. 

Then,  taking  a  decimeter  (one  tenth  of   a 

meter)  as  a  new  unit,  suppose  this  is  contained    ^  5  S  2  S 

three  times  in  A^B  with  a  remainder  AJi  less  ^i        ^2-B 

than  a  decimeter.  * 

Finally,  using  as  a  unit  a  centimeter  (one  tenth  of  a  decimeter),  sup- 
pose this  is  contained  exactly  six  times  in  A^B. 

Then,  the  length  of  ^^  is  5  meters,  3  decimeters,  and  6  centi- 
meters, or  5.36  meters. 

The  process  of  measuring  considered  here  is  ideal.  In  practice  we 
cannot  say  that  a  given  segment  is  contained  exactly  an  integral 
number  of  times  in  another  segment.     See  §  235. 

112 


MEASUEEMENT  OF  LINE-SEGMENTS.  113 

234.  It  may  also  happen  that,  in  continuing  this  ideal 
process  of  measuring  as  just  described,  no  subdivided 
unit  can  be  found  which  exactly  measures  the  last  interval, 
that  is,  such  that  the  final  division  point  falls  exactly 
on  B. 

E.g.  it  is  known  that  in  a  square  whose  sides  are  each  one  unit 
the  diagonal  is  V'2,  and  that  this  cannot  be  exactly  expressed  as  an 
integer  or  a  fraction  whose  numerator  and  denomi- 
nator are  both  integers. 

By  the  ordinary  process  of  extracting  square  root 
we  find  V2  =  1.4142  •••,  each  added  decimal  mak- 
ing a  nearer  approximation.  But  this  process  never 
terminates. 

Hence,  in  attempting  to  measure  the  diagonal  of  a  square  whose 
side  is  one  meter,  we  find  1  meter,  4  decimeters,  1  centimeter,  4  milli- 
meters, etc.,  or  1.414  meters  approximately. 

It  should  be  noticed,  however,  that  1.415  is  greater  than  the  di- 
agonal and  hence  the  approximation  given  is  correct  within  one 
millimeter. 

235.  Evidently  any  line-segment  can  be  measured  either 
exactly  or  to  a  degree  of  approximation,  depending  upon 
the  fineness  of  the  instruments  and  the  skill  of  the  opera- 
tor. The  word  measure  is  commonly  used  to  include 
both  exact  and  approximate  measurement. 

For  practical  purposes,  a  line-segment  is  measured  as 
soon  as  the  last  remainder  is  smaller  than  the  smallest 
unit  available.  It  should  be  noticed  that  all  practical 
measurements  are  in  reality  only  approximations,  since  it 
is  quite  impossible  to  say  that  a  given  distance  is,  for  in- 
stance, exactly  25  ft.  It  may  be  a  fraction  of  an  inch 
more  or  less. 

E.g.  in  the  above  example  1.414  meters  gives  the  length  of  the 
diagonal  for  practical  purposes  if  the  millimeter  is  the  smallest  unit 
available.     The  error  in  this  case  is  less  than  one  millimeter. 


114  PLANE  GEOMETRY. 

236.  Definition.  Two  straight  line-segments  are  com- 
mensurable if  they  have  a  common  unit  of  measure.  Other- 
wise they  are  incommensurable. 

E.g.  two  hne-segments  whose  lengths  are  exactly  5.27  and  3.42 
meters  respectively  have  one  centimeter  as  a  common  unit  of  meas- 
ure, it  being  contained  527  times  in  the  first  segment  and  342  times 
in  the  second. 

But  the  side  and  the  diagonal  of  a  square  have  no  common  unit  of 
measure. 

In  the  example  of  §  234,  the  millimeter  is  contained  1000  times  in 
the  side  and  1414  times  in  the  diagonal,  plus  a  remainder  less  than 
one  millimeter.  A  similar  statement  holds  for  any  unit  of  measure, 
however  small. 

237.  For  the  purposes  of  practical  measurement  any  two 
line-segments  mai/  be  considered  as  commensurable,  but  for 
theoretical  purposes  it  is  necessary  to  take  account  of  in- 
commensurable segments  also. 

The  theorems  in  this  chapter  are  here  proved  for  com- 
mensurable segments  only.  They  are  proved  for  incom-- 
mensurable  segments  also  in  Chapter  VII. 

RATIOS  OF   LINE-SEGMENTS. 

238.  The  ratio  of  two  commensurable  line-segments  is  the 
quotient  of  their  numerical  measures  taken  with  respect  to 
the  same  unit. 

E.g.  if  two  segments  are  respectively  3  ft.  and  4  ft.  in  length,  the 
ratio  of  the  first  segment  to  the  second  is  |  and  the  ratio  of  the 
second  to  the  first  is  f. 

239.  The  ratio  of  two  commensurable  segments  is  the 
same,  no  matter  what  common  unit  of  measure  is  used. 

E.g.  two  segments  whose  numerical  measures  are  3  and  4  if  one 
foot  is  the  common  unit,  have  36  and  48  as  their  numerical  measures 
if  one  inch  is  the  common  unit.  But  the  ratio  is  the  same  in  both 
cases,  namely  :  ||  =  |. 


MEASUREMENT  OF  LINE-SEGMENTS.  115 

240.  The  approximate  ratio  of  two  incommensurable  line- 
segments  is  the  quotient  of  their  approximate  numerical 
measures.  It  will  be  seen  that  this  approximate  ratio 
depends  upon  the  length  of  the  smallest  measuring  unit 
available,  and  that  the  approximation  can  be  made  as 
close  as  we  please  by  taking  the  measuring  unit  small 
enough. 

E.g.  an  approximate  ratio  of  the  side  of  a  square  to  its  diagonal 

is = Another  and  closer  approximation  is = ■.    In 

1.41     141  1.414     1414 

this  case  the  numerical  measure  of  one  of  the  segments  is  exact. 

An  approximate  ratio  of  V2  to  VS,  in  which  neither  has  an  exact 

1 41      141 

measure,  is  — —  =  — .     Another  is 


1.73      173  1.732      1732 

241.  It  should  be  clearly  understood  that  the  numerical 
measure  of  a  line-segment  is  a  number,  as  is  also  the  ratio 
of  two  such  segments.  Hence  they  are  subject  to  the 
same  laws  of  operation  as  other  arithmetic  numbers. 

For  example,  the  following  are  axioms  pertaining  to 
such  numbers  : 

(1)  Numbers  which  are  equal  to  the  same  number  are 
equal  to  each  other. 

(2)  If  equal  numbers  are  added  to  or  subtracted  from 
equal  numbers^  the  results  are  equal  numbers. 

(3)  If  equal  numbers  are  multiplied  by  or  divided  by 
equal  numbers.,  the  results  are  equal  numbers. 

It  is  understood,  however,  that  all  the  numbers  here 
considered  are  positive.  For  a  more  complete  con- 
sideration of  axioms  pertaining  to  numbers,  see  Chapter 
I  of  the  Advanced  Course  of  the  authors'  High  School 
Algebra. 


116  PLANE  GEOMETRY. 

242.    A  proportion  is  an  equality,  each  member  of  which 
is  a  ratio.     Four  numbers,  a,  5,  c^  and  c?,  are  said  to  be  in 

proportion,  in  the  order  given,  if  the  ratios  —  and  -   are 

0  d 

equal.  In  this  case  a  and  c  are  called  the  antecedents 
and  h  and  d  the  consequents.  Also  a  and  d  are  called 
the  extremes  and  h  and  c  the  means. 

The  proportion  -  =  —  is  sometimes  written  a  :  b  =  c  :  d, 
0      d 

and  in  either  case  may  be  read  a  is  to  b  as  e  is  to  d. 

If  D  and  E  are  points  on  the  sides  of  the  triangle  ABC^ 
and  if  w,  n,  jt?,  and  q^  the  numerical  measures  respectively 

of  AD,  DB,  AE,  and  EC,  are  such  that  —=E^ 

n      q 

then  the  points  D  and  E  are  said  to  divide 

the  sides  AB  and  AC  proportionally,  that  is, 

in  the   same   ratio.     For   conveftience   it  is 

common  to  let  AB,  DB,  AE,  and  EC  stand  for 

the  numerical  measures  of  these  segments,   and  thus  to 

AD      AE 
write  the  above  proportion,  —  =  —  or  AD  :  DB  =  AE:EC. 

DB      EC 


THEOREMS  ON  PROPORTIONAL  SEGMENTS. 

243.  Theorem.  If  a  line  is  parallel  to  one  side  of  a 
triangle  and  cuts  the  other  two  sides,  then  it  divides  these 
sides  in  the  same  ratio. 

Given  A  ABC  in  which  DE  II BC. 

To  prove  that— =— . 

DB       EC 
Proof  :    Choose  some  common  measure 
of  AD  and  DB,  as  AK.     Suppose  it  is  con- 
tained 3  times  in  AD  and  5  times  in  DB. 


MEASUREMENT  OF  LINE-SEGMENTS. 


117 


Then 


AD 


(1) 


Through  the  points  of  division  on  AD  and  DB  draw  lines 
parallel  to  5C,  cutting  AE  and  EC.  By  §  155  these  paral- 
lels divide  AE  into  three  equal  parts  and  EC  into  five 

equal  parts.     Hence,         ^^      ^ 

(2) 


EC 


=h 


§241 


.-.  from  (1)  and  (2)  ^  =  ^. 
^       DB       EC 

For  a  proof  in  case  AD  and  DB  are  incommensurable,  see  §  410. 

244.  EXERCISES. 

1.   If  DEW  EC  in  A  ABC,  compute  the  segments 
left  blank  from  those  given  in  the  following  table : 


AD 

DB 

AE 

EC 

AB 

AC 

20 

24 

15 

4 

56 

42 

102 

12 

408 

25 

18 

342 

Required  to  com- 


2.   5  is  a  point  visible  from  A  but  inaccessible, 
pute  the  distance  from  A  to  B. 

Suggestion.  Select  some  accessible  point  C 
from  which  A  and  B  are  both  visible.  Through  E, 
a  point  near  A  and  on  the  line  of  sight  from  A  to 
C,  draw  ED  II  CB  and  meeting  the  line  of  sight 
from  ^  to  5  at  D. 

Now  AE,  EC,  and  AD  can  be  measured.     Then     _ 
i)i?  and  hence  ^5  can  be  computed  by  §  243. 

Note.  The  advantage  of  this  method  over  that  in 
§  34,  Ex.  5,,  is  that  here  a  small  triangle  AED  is  made  to 
do  the  service,  which  was  there  performed  by  another 
triangle  the  same  size  as  ABC. 


i± 


i:Li. 


118  PLANE  GEOMETRY. 

245.   Theorem.     If  four  numbers  m,  n,  p,  q  are  such 

that  —  =  -,  then  it  follows  that: 
n      q 


<"S= 

1. 
P 

(2)  ^  =  ^. 

(3)  '»  + 

n 

—  =1 

p  +  q                         (^)m-n      p-q^ 
q                                       n             q 
(5)  m-hn_p  +  q 
m  —  n      p  —  q 

Given 

n      q 

To  prove  (1)  divide  the  members  of  1  =  1  by  those  of 

To  prove  (2)  multiply  each  member  of  (a)  by  -• 

P 
To  prove  (3)  add  1  to  each  member  of  (a)  and  reduce 

each  side  to  a  common  denominator. 

To  prove  (4)  subtract  1  from  each  member  of  (a)  and 
reduce  each  side  to  a  common  denominator. 

To  prove  (5)  divide  the  members  of  (3)  by  the  members 
of  (4). 

Write  out  these  proofs  in  full,  giving  the  reason  for 
each  step,  and  read  off  the  results  as  applied  to  the  figure. 

For  example,  show  that  (3)  gives,  when  applied  ^^ 

to  the  figure,  _   >^    V 

AB^AC  n^'^^q 

DB       EC  /- 1 

246.  The  results  in  the  above  theorem  are  sometimes 
named  as  follows : 

The  proportion  (a)  is  said  to  be  taken  by  inversion  in 
(1),  by  alternation  in  (2),  by  composition  in  (3),  by  division 
in  (4),  by  composition  and  division  in  (5). 


MEASUREMENT  OF  LINE-SEGMENTS. 


119 


247. 


EXERCISES. 


1.  If  —  =^,  prove  that =  —^ — ,  aud  hence    show    in  the 

n       q  m  -\-  n      p  +  q 

above  figure  that  AD      AE 

AB^^A^' 

2.  If  in  the  figure  on  page  118  DEWBC,  compute  the  segments 
indicated  by  blanks  in  the  accompanying  table. 


3. 

If     ^ 

n 

9 

show 

that^  +  '^  = 
P 

n  +  q 

4. 

If     ^ 
n 

show  that  '"^-P  = 
n  +  q 

m 

n ' 

^       5. 

If  ^  = 

X 

^   and 
7 

m_p 

y    q 

show 

that  a: 

=  y- 

\    6. 

If  !^  = 

*    X 

=  -,  show  that 

y 

x  =  y 

7.    State  Exs.  4,  5,  and  6 
in  words. 

AD 

AB 

DB 

AE 

AC 

EC 

8 

12 

6 

6 

10 

16 

6 

9 

7 

12 

10 

.10 

10 

8 

18 

10 

7 

14 

240 

200 

380 

160 

140 

20 

120 

100 

50 

35 

21 

14 

40 

15 

30 

500 

200 

400 

90 

^4% 

70  „ 

20 

.30 

^   - 

800 

360 

300 

27 

30 

48 

30 

20 

50 

27 

560 

48 

8.   A  triangle  is  formed  by  a  chord  and  the  tangents  to  the  circle 
at  its  extremities.     Prove  that  the  triangle  is  isosceles. 
|k       9.    A  triangle  with  angles  A,  B,  C  is  circumscribed  about  a  circle. 
Find  the  angles  of  the  triangle  formed  by  the  chords  joining  the 
points  of  tangency. 


120 


PLANE  GEOMETRY. 


248.  Theorem.     //  a  line  divides  two  sides  of-  a  tri- 
angle in  the  same  ratio,  it  is  parallel  to  the  third  side. 
Given  the  points  D  and  E  on  the  sides  of  the  A  ABC  such 

,^  ,  AD     AE 
that  —  =  — . 
DB      EC 

To  prove   that  BE  W  BC. 

Proof :  Suppose  de'  is  drawn  paral- 
lel to  BC.  It  is  proposed  to  prove  that 
the  point  E^  coincides  with  E. 

AB      AC 


Since  BE'  II  BC,  we  have 


AB      E'C 


But  by  (3),  §  245,  ^  =  ^. 
-    ^   ^    ^  DB       EC 

Now   use   the   proof   of    Ex.   5,    §  247,   to    show   that 

e'c  =  EC^    and    hence    that   e'    and    E   coincide,    so    that 

BE  II  BC.     Give  the  proof  in  full  detail. 


249. 


EXERCISES. 


1.  Show'  by  §  248  that  the  Hne  joining  the  middle  points  of  two 
sides  of  a  triangle  is  parallel  to  the  third  side.     Compare  §  151. 

2.  Show  by  §  243  that  the  line  which  bisects  one  side  of  a  triangle 
and  is  parallel  to  a  second  side  bisects  the  third  side. 

3.  If  in  the  A  ABC  a  segment  Cy  connects  the 
vertex  to  any  point  y  of  the  base,  find  the  locus  of 
the  point  x  on  this  segment  such  that  Cx  :  Cy  is 
the  same  for  all  points  y. 

4.  ABCD  is  a  O  whose  diagonals  meet  in  0. 
If  ?/  is  a  point  on  any  side  of  the  O,  find  the  locus 
of  a  point  x  on  the  segment  Oy  such  that  Oy  :  xy 
is  the  same  for  every  such  point  y. 

5.  Find  the  locus  of  the  points  of  intersection 
of  the  medians  of  all  triangles  having  the  same 
base  and  equal  altitudes.     (Use  §§  158,  248.) 


MEASUBEMENT  OF  LINE-SEGMENTS. 


121 


250.  Theorem.  The  bisector  of  an  angle  of  a  triangle 
divides  the  opposite  side  into  segments  whose  ratio  is  the 
same  as  that  of  the  adjacent  sides. 

Given  CD  bisecting  Z  C  in  A  ABC, 

r^  ...    AD      AC 

To  prove  that  —  =  — . 

DB       BC 
Proof :     Through  B  draw  BE  II  DC. 
Prolong  ^C  to  meet  BE  at  E. 

In  A  ABE        ^^4£,    (Why?) 
DB       CE      ^         '^    ^ 

Now  show  that  A  BCE  is  isosceles,  and  hence  that  BG 
may  be  substituted  for  CE  in  the  above  proportion.  Com- 
plete the  proof. 


251. 


EXERCISES. 


1.   Fill  in  the  blank  spaces  in  the  table,  if  in  the  figure  of  §  250 
CD  is  the  bisector  of  Z  ^  CB. 


AC 

CB 

AD 

DB 

AB 

8 

10 

6 

20 

16 

12 

35 

17 

40 

3 

2 

1 

7 

9 

12 

12i 

8 

16 

364 

200 

480 

54 

65 

105 

24.5 

18.3 

32.6 

122  PLANE  GEOMETUY. 

252.  Definition.  A  segment  is  said  to  be  divided  exter- 
nally by  any  point  which  lies  on  the  line  of  the  segment  but 
not  on  the  segment  itself. 

E.g.  point  C  divides  the  segment  AB  externally,  the  parts  being 
^ C  and  CB,  while  point  B  divides  the  segment  AC  in- 
ternally,  the  parts  being  ^5  and  BC.  •       '        ' 

253.'  Theokem.  a  line  which  bisects  an  exterior 
angle  of  a  triangle  divides  the  opposite  side  externally 
into  two  segments  whose  ratio  is  the  same  as  that  of  the 
adjacent  sides  of  the  triangle. 


Given  CD,  the  bisector  of  the  exterior  angle  at  C  of  the  triangle 
ABC. 

_,  .,     ,  AD      AC 

To  prove  that  —  =  — . 

BD      BC 
Proof :     Through  B  draw  BE  II  DC. 

In  A  ACD  AD^AC  (Why?) 

BD       EC  \        J    y 

Now   show   that   A  EBC  is    isosceles,   and   hence   that 

EC  =  BC. 
Complete  the  proof. 

254.  EXERCISES. 

1.  Draw  a  triangle  with  an  acute  exterior  angle  bisected.     Using 
different  lettering  from  that  in  §  253,  prove  the  theorem  again. 

2.  Compare  the  proofs  in  §§  250  and  253.     Give  the  proof  in  §  253 
for  a  figure  in  which  AC  <  BC. 


MEASUREMENT  OF  LINE-SEGMENTS. 

3.   Fill  in  the  blank  spaces  in  the  table  below  if 
CD  is  the  bisector  of  the  exterior  angle  BCK. 


123 


AC 

CB 

AD 

DB 

AB 

7 

4 

9 

14.3 

9.6 

18 

164 

48 

144 

13.7 

84 

60 

32 

60 

25 

56 

80 

40 

4.8 

12.5 

6 

550 

600 

400 

350 

200 

300 

c- 


/  2 


^K 


4.   To  measure  indirectly  the  distance  from  an  accessible  point  A 
to  an  inaccessible  point  B  by  means  of  §  253. 

Suggestion.  Through  C,  a  point  where  A 
and  B  are  both  visible,  draw  CK  making  Z 1  =  Z  2. 
Produce  KC  to  a  point  D  on  the  line  BA  extended.   •^' 


A  B 

What  lines  must  now  be  measured  in  order  to  compute  AB1 

5.  What  methods  have  been  used  so  far  for  the  indirect  measure- 
ment of  the  distance  from  an  accessible  to  an  inaccessible  point? 
Compare  these 

(a)  As  to  the  simplicity  of  the  theory  involved. 
(5)  As  to  the  simplicity  and  ease  of  the  direct  measurements  re- 
quired. 

6.  Divide  a  given  line-segment  in  a  given  ratio  without  construct- 
ing a  line  parallel  to  another. 

7.  Similarly  divide  a  given  line-segment  externally  in  a  given  ratio. 

8.  Solve  Ex.  3,  §  249,  if  x  is  on  Cy  extended. 

9.  Solve  Ex.  4,  §  249,  if  x  is  on  Oy  extended. 


124  PLANE  GEOMETRY. 


SIMILAR  POLYGONS. 


255.  Two  polygons,  in  which  the  angles  of  the  one  are 
equal  respectively  to  the  angles  of  the  other,  taken  in 
order,  are  said  to  be  mutually  equiangular. 

The  angles  of  the  two  polygons  are  thus  arranged  in 
pairs  of  equal  angles,  which  are  called  corresponding 
angles. 

Two  sides,  one  of  each  polygon,  included  between  cor- 
responding angles,  are  called  corresponding  sides. 

256.  Two  polygons  are  similar  if  (1)  they  are  mutually 
equiangular  and  if  (2)  their  pairs  of  corresponding  sides 
are  proportional. 

Two  polygons  may  have  property  (1)  but  not  (2).  For  example,  a 
rectangle  and  a  square.  Or  they  may  have  property  (2)  and  not  (1). 
For  example,  a  square  and  a  rhombus. 

Hence  any  proof  that  two  polygons  are  similar  must  show  that 
both  (1)  and  (2)  hold  concerning  them. 

In  the  case  of  triangles  it  will  be  proved  that  either  property  speci- 
fied in  the  definition  of  similar  polygons  is  sufficient  to  make  them 
similar. 

257.  Theorem.  If  two  triangles  are  mutually  equi- 
angular, they  are  similar. 


B  c 


Given  A  ABC  and  A'B'C,  in  which  Z  A  =  Z.A',  ZB=  Z  B', 
and  ZC  =  ZC'. 


MEASUREMENT  OF  LINE-SEGMENTS.  125 

To  prove  that  the  other  property  of  similarity   holds, 

1      ..     .    AB         BC         CA 

namely  that = =  — — ,. 

*"  A'B'      B'C'      c'a' 

Proof:     Place  A  a'b'c'  on  A  ABC  with  Z  A'  upon  its 
equal  Z  A,  and  b'c'  taking  the  position  BE. 

Now  show  that  BE  II  BC  and  hence  —  =— , 

, n    ,  .  »  AB        AC 

that  IS, 


^'B'      A'& 
In  like  manner,  placing  Z  ^'  upon  Z  B, 

show  that  4^  =  4^. 

Give  the  full  details  of  this  proof. 


258.  EXERCISES. 

1.    To  measure  indirectly  the  distance  from  an  accessible  point  A 
to  an  inaccessible  point  B. 


from  A  to  B,  and  ED  ±  AD.     Let  C  be  the  point  -27----iL 
on  AD  which  lies  in  line  with  E  and  B. 

Now  show  that  S^EDC  and  BAG  are  mutually  equiangular  and 
hence  similar.  What  segments  need  to  be  measured  in  order  to  com- 
pute AB'i    Give  full  details  of  proof. 

2.  Prove  that  two  right  triangles  are  similar  if  they  have  an  acute 
angle  of  one  equal  to  an  acute  angle  of  the  other. 

3.  Two   isosceles   triangles  are  similar  if  they  have  the  vertical 
angle  of  one  equal  to  the  vertical  angle  of  the  other. 

4.  Two  triangles  which  have  the  sides  of  one  respectively  parallel 
or  perpendicular  to  the  sides  of  the  other  are  similar. 

5.  Show  by  similar  triangles  that  the  segment  joining  the  mid- 
points of  two  sides  of  a  triangle  is  equal  to  one  half  the  third  side. 


126 


PLANE  GEOMETRY. 


259.  Theorem.  //  two  triangles  have  an  angle  of  one 
equal  to  an  angle  of  the  other  and  the  'pairs  of  adjacent 
sides  in  the  same  ratio,  the  triangles  are  similar. 


Z  A'  and 


AB 
A'B' 


AC 

A'C' 


Given  A  ABC  and  A'B'C  in  which  ZA 

To  prove  that  A  ABC'^AA'b'c'. 

Proof:    Place  AA'b'c'  upon  A  ABC  with  ZA'  on  Z  A, 
B'c'  taking  the  position  BE. 

Then,  AB^AC  (Why?) 

AD      AE  V        J    y 

and  hence,  BE  \\  BC.  (Why  ?) 

Now  show  that  A  ABE  and  ABC  are  mutually  equian- 
gular and  hence  similar. 

260.  Theorem.    //  two  tri- 
angles   have   their  pairs   of 
corresponding    sides    in    the 
same  ratio,  they  are  similar. 
Given   A  ABC   and    A'B'C    in 
AB  ^  BC  ^  CA 
A'B'      B'C      CA' ' 
B  c  To     prove     that     Aabc^^ 

A  A' b'c',  that  is,  to  prove  Z^  =  Z  ^',  Zb  =  Z  b\  Z  c=Zg'. 


C    which 


MEASUREMENT  OF  LINE-SEGMENTS.  127 

Proof :    Lay  off  on  AB  and  AC  respectively  AD  =  A'b' 
and  AE  =  A'c',  and  draw  DE. 

Now  prove,  as.  in  §  259,  that  AADE^AABC^ 

and  hence,  that  —  = (1^ 

DE      BC  ^  ^ 

B"t'  i^  =  ~-     (Why?)  (2) 

Hence,  since  AD  =  A'b',  it  follows  from  (1)  and  (2)  that 
de=b'c',  as  in  Ex.  5,  §  247. 

Now  show  that      A  A'b'c'^  A  ADE, 

and  hence,  that  AA^b'^-^AABC. 

Make  an  outline  of  the  steps  in  this  proof  and  show  how  each  is 
needed  for  the  one  that  follows. 


261.  EXERCISES. 

1.  Given  a  triangle  whose  sides  are  2,  3,  4.  Construct  a  triangle 
having  its  angles  equal  respectively  to  those  of  the  given  triangle  and 
having  a  side  10  corresponding  to  the  given  side  2. 

2.  If  each  of  two  triangles  is  similar  to  a  third  triangle,  they  are 
similar  to  each  other. 

3.  If  in  a  right  triangle  a  perpendicular  be  drawn  from  the  vertex 
of  the  right  angle  to  the  hypotenuse,  show  that  each  of  the  triangles 
thus  formed  is  similar  to  the  given  tri- 
angle, and  hence  that  they  are  similar  to 
each  other. 


4.  In  the  figure  of  Ex.  3,  make  a  table 
showing  which  angles  are  equal  and  which 
pairs  of  sides  are  corresponding  in  the  fol- 
lowing pairs  of  triangles:  A  CD  and  ACB,  CDB  and  ACB,  A  CD 
and  CDB. 

5.  On  a  given  segment  as  a  side  show  how  to  construct  a  triangle 
similar  to  a  given  equilateral  triangle. 


128  PLANE  GEOMETRY. 

262.  Theorem.  The  square  on  the  hypotenuse  of  a 
right  triangle  is  equal  to  the  sum  of  the  squares  on  the 
other  two  sides.  ^       . .         ^^"^^^'u        \ 


^Ia 


I 

Given  A  ABC  with  a  right  angle  at  C.  Call  the  lengths  of  the 
sides  opposite  A  A,  B,  C,  respectively,  a,  b,  c. 

To  prove  that  c^  =  a^  -{-  h"^. 

Proof :  Let  the  perpendicular  p  divide  the  hypotenuse 
into  tlie  two  parts  m  and  n  so  that  c  =  m  -\-  n. 

From  A  ACB  and  ACB  show  that  ^  =  -  (1) 

h       c 

From  A  CDB  and  ACB  show  that  -  =  -•  (2) 

From  (1)  mc  =  lP'  ^      "^  (3) 

and  from  (2)  nc  =  a^     (Why?).  (4) 

From  (3)  and  (4)  (m  +  n)c  =  0^  +  1)^       (Why?). 
That  is,  c  •  c  =  e^  =^  a?" -\-  h^. 
For  another  proof  of  this  theorem  see  §  319. 

Historical  Note.  The  proof  given  above  is  supposed  to  be 
that  given  by  Pythagoras,  who  first  discovered  the  theorem. 

263.  EXERCISES. 

1.  The  radius  of  a  circle  is  8.  "What  is  the  distance  from  the 
center  to  a  chord  v^hose  length  is  6  ? 

2.  In  the  same  circle,  what  is  the  length  of  a  chord  whose  distance 
from  the  center  is  5  ? 

3.  Find  the  diagonal  of  a  square  whose  side  is  5 ;  whose  side  is  a. 

4.  What  is  the  side  of  a  square  whose  diagonal  is  8  ?  whose  diagonal 
isrf? 


i.„vl 


V 


MEASUREMENT  OF  LINE-SEGMENTS. 


129 


5.  The  hypotenuse  of  a  right  isosceles  triangle  is  12  inches.     Find 
the  lengths  of  its  sides. 

6.  The  diagonals  of  a  rhombus  are  14  and  10  inches  respectively. 
Find  the  length  of  its  sides.     (See  §  147,  Ex.  4.) 

C»     7.    The  square  on  the  hypotenuse  of  a  right  triangle  is  equal  to  four 
times  the  square  on  the  median  to  the  hypotenuse. 

8.  What  is  the  radius  of  a  circle  if  a  chord  12  inches  long  is  9 
inches  from  the  centre  ? 

9.  Find  the  altitude  of  an  equilateral  triangle  whose  side  is  8 ; 
whose  side  is  a. 

10.   If  the  altitude  of  an  equilateral  triangle  is  h,  find  its  side. 
(Use  the  formula  obtained  under  Ex.  9.) 
'  f    11.   Find  the  altitude  of  a  triangle  whose  sides  are  6,  8,  and  10. 

12.  The  oval  in  the  figure  is  a  design  used  in  the  construction  of 
sewers.  It  is  constructed  as  follows : 
In  the  O  OA  let  CD,  the  perpendicular 
bisector  of  AB,  meet  the  arc  AO'B  at  0'. 
Arcs  AM  and  BN  are  drawn  with  the 
same  radius  AB  and  with  centers  B  and 
A  respectively. 

The  lines  BO'  and  A  0'  meet  these  arcs 
in  M  and  N  respectively. 

The  arc  MDN  has  the  center  0'  and 
radius  O'M. 

(a)    Is  arc  ACB  tangent  to  AM  and 
BN  at  A  and  B  respectively  ?     Why  ? 

Q))    Is  arc  MDN  tangent  to  AJl  and 
BN  at  M  and  N  respectively  ?     Why  ? 

(c)    If  ^5  =  8  feet,  find  BO',  and  hence,  O'M,  and  finally  CD. 
That  is,  if  the  sewer  is  8  feet  wide,  what  is  its  depth? 

(c?)   If  the  width  of  the  sewer  is  a  feet,  show  that  its  depth  is 

I  (4  -  V2). 

(e)    If  the  depth  of  the  sewer  is  d  feet,  show  that  its  width  is 
c?  (4  +  V2) 
7 

(/)   Compute  to  two  places  of  decimals  the  width  of  a  sewer  whose 
depth  is  12  feet. 


130  PLANE  GEOMETRY. 

264.  Theorem.  //  in  two  right  triangles  the  hypote- 
nuse of  the  one  equals  the  hypotenuse  of  the  other,  and  if 
the  sides  a,  h  and  a',  ¥  are  such  that  a  >  a\  then  h  <  b\ 

Proof  :    Let  c  be  the  length  of  the  hypotenuse  in  each. 

Then  ^2  +  52^  ^2  ^nd  a'^+b''^=  A  (Why?) 

...  ^2 +  52 ^^^2 ^5^2.  (Why?) 

and  a2  _  a'^  =  h'^  -  h\  (Why  ?)  «> 

Since  a>a\  the  left  member  of  the 
last  equation  is  positive,  and  hence  the 
right  member  is  also  positive,  that  is,  h<b'. 

265.  Theorem.  In  the  same  circle  or  in  equal  circles, 
of  two  unequal  chords,  the  greater  is  nearer  the  center. 


Given  (D  OA  and  O'A'  in  which  OA  =  O'A'  and  AB>A'B'. 
To  prove  that  AB  is  nearer  the  center  than  A'b'. 
Proof :  Draw  the  J^  b  and  b'  and  the  radii  c  and  c'. 
Then  a  and  a'  are  halves  of  AB  and  A'b'  respectively. 
Now  complete  the  proof,  using  §  264. 

266.  Theorem.     State  and  prove  the  converse  of  the 
theorem  in  §  265,  using  the  same  figure. 

267.  Definitions.     A  continued  proportion  is  a  series  of 
equal  ratios  connected  by  signs  of  equality. 

^•^'    b-d-f-h 
The  perimeter  of  a  polygon  is  the  sum  of  its  sides. 


MEASUREMENT  OF  LINE-SEGMENTS.  131 

268.  Theoeem.  In  a  continued  proportion,  the  sum 

of  the  antecedents  and  the  sum  of  the  consequents  form  a 

ratio  equal  to  any  one  of  the  given  ratios. 

a.     c      c      Q 
Given  the  continued  proportion  -  =  -  =  -  =  ^ . 

b     d     f     h 

To  prove  that  ,       ,      ,^  =  -  • 

h-Vd+f+h      h 

Proof  :     Let  T  =  ^• 

0 

™,  c  e  q 

Then,  ^  =  .,    j  =  r,    j  =  r. 

! 

Hence,  a  =  hr,    e  =  dr^    e=fr^   g  =  hr^ 

and    a-\-c  +  e-\-g  =  hr-^dr  -\-fr  +  hr  =  (7>  +  d  +/+  h)r, 

or  — ' ■ -^  z=r  =  —• 

h+d+f+h  b 

Give  all  reasons  in  full. 

269.  Theorem.  The  perimeters  of  two  similar  poly- 
gons are  in  the  same  ratio  as  any  two  corresponding 
sides. 


D 

Proof  :     By  definition  of  similar  polygons 

ab  _  bc  _  cd  _  be  __  ea 

a'b'~  b'c'~~  c'b'~  d'e' ~~  e'a' 

Complete  the  proof. 


132  PLANE  GEOMETRY. 

270.   Theorem.     //  the  diagonals  drawn  from  one 
vertex  in  each  of  two  polygons  divide  them  into  the  same 
number  of  triangles,  similar  each  to  each  and  similarly 
placed,  then  the  two  polygons  are  similar, 
B  ^ 


Given  the  diagonals  drawn  from  the  vertices  A  and  A'  in  the 
polygons  P  and  P',  forming  the  same  number  of  triangles  in  each, 
such  that     AI^AI',   All ^  All',   AIII'^AIII'. 

To  prove  that  P  -^  p'. 

Outline  of  Proof  :     (1)  Use  the  hypothesis  to  show  that 

Za  =  Za',^b  =  Zb\   Zc=Zc',  etc. 

(2)  Show  that       4^  =  4^  =  44^  etc. 
^  ^  a'b'     b'c'     c'd' 

TiC  CT) 

Notice  that  the  proportion    ——-  =  ——-    follows   from 

b'c'     c'd' 

¥c'~A^~C^''  ^^'^^^ 

Give  the  proof  in  detail. 

271.  Theorem.  State  the  converse  of  the  "preceding 
theorem,  and  give  the  proof  in  full  detail. 

Make  an  outline  of  all  the  steps  in  the  proofs  of  these 
two  theorems. 


MEASUREMENT  OF  LINE-SEGMENTS.  133 

272.  Theokem.  //  through  a  fixed  point  within  a 
circle  any  number  of  chords  are  drawn,  the  product  of  the 
segments  of  one  chord  is  equal  to  the  product  of  the  seg- 
ments of  any  other. 


Given  O  C  with  any  two  chords  AB  and  DE  intersecting  in  P. 

To  prove  that        AP  ■  PB  =  ep  •  PB. 
Proof  :     Draw  EB  and  DA. 

TheD,  Z1  =  Z2  andZ3  =  Z4.  (Why?) 

Hence,  A  EPB  ~  A  PDA,  (Why?) 

Which  are  corresponding  angles  and  which  are  corre- 
sponding sides? 

Show  that  — :  = 

EP       PA 
Complete  the  proof. 

It  follows  from  this  theorem  that  if  a  chord  AB  is  made 

to  swing  around  the  fixed  point  P,  the  product  AP  •  PB 

does  not  change,  that  is,  it  is  constant. 

273.  EXERCISES. 

1.  Which  chord  through  a  point  is  bisected  by  the  diameter  through 
that  point?     Why? 

^y/"  2.   Through  a  given  point  within  a  circle  which  chord  is  the  short- 
Zest?    Why? 

3.  The  product  AP-  PB  is  the  area  of  the  rectangle  whose  base 
and  altitude  are  the  segments  of  AB.     (See  §  307.) 

Note  that  this  area  is  constant  as  the  chord  swings  about  the  point 
P  as  a  pivot. 


134  PLANE  GEOMETRY. 

274.  Definition.  If  a  secant  of  a  circle  is  drawn  from  a 
point  P  without  it,  meeting  the  circle  in  the  points  A  and 
J3,  then  PB  is  called  the  whole  secant  and  PA  the  external 
segment,  provided  A  lies  between  B  and  P. 

275.  Theoeem.  If  from  a  fixed  point  outside  a  cir- 
cle any  number  of  secants  are  drawn,  the  product  of  one 
whole  secant  and  its  external  segment  is  the  same  as  that 
of  any  whole  secant  and  its  external  segment, 

'P 


Given  secants  PB  and  PE  drawn  from  a  point  P. 

To  prove  that  PA-  PB  —  PD-  PE. 

Proof  :     In  the  figure  show  that  A  PDB  ~  A  PAE. 

Complete  the  proof. 

276.  EXERCISES. 

1.  A  point  P  is  8  inches  from  the  center  of  a  circle  whose  radius 
is  4.  Any  secant  is  drawn  from  P,  cutting  the  circle.  Find  the 
product  of  the  whole  secant  and  its  external  segment. 

2.  From  the  same  point  without  a  circle  two  secants  are  drawn. 
If  one  whole  secant  and  its  external  segment  are  14  and  5  respec- 
tively and  the  other  external  segment  is  7,  find  the  other  whole 
secant. 

3.  Two  chords  intersect  within  a  circle.  The  segments  of  one  are 
m  and  n  and  one  segment  of  the  other  is  p.  Find  the  remaining 
segment. 


MEASUREMENT  OF  LINE-SEGMENTS.  135 

277.  Theorem.  If  a  tangent  and  a  secant  meet  out- 
side a  circle,  the  square  on  the  tangent  is  equal  to  the 
product  of  the  whole  secant  and 
its  external  segment. 

Proof :     Show  that  /    /         ^^^ 

A  APD  ~  A  BPD  and  hence  that 
PB  :  PB  —  PD  :  PA.  B^ 

Complete  the  proof. 

278.  EXERCISES. 

1.  If  a  square  is  constructed  on  PD  as  a  side,  and  a  rectangle  with 
PB  as  base  and  PA  as  altitude,  compare  their  areas  as  the  secant 
revolves  about  P  as  a  pivot. 

2.  Show  that  the  theorem  in  §  277  may  be  obtained  as  a  direct 
consequence  of  that  in  §  275  by  supposing  one  secant  to  swing  about 
P  as  a  pivot  till  it  becomes  a  tangent. 

3.  A  point  P  is  10  inches  from  the  center  of  a  circle  whose  radius 
is  6  inches.     Find  the  length  of  the  tangent  from  P  to  the  circle. 

4.  The  length  of  a  tangent  from  P  to  a  circle  is  7  inches,  and  the 
external  segment  of  a  secant  is  4  inches.  Find  the  length  of  the 
whole  secant. 

5.  What  theorems  are  included  in  the  following  statement  : 
"  Froin  a  point  P  in  a  plane  a  line  is  drawn  cutting  a  circle  in  A  and  B. 
Then  the  product  PA  -  PB  is  the  same  for  all  such  lines  "  ? 

6.  In  a  circle  of  radius  10  a  point  P  divides  a  chord  into  two  seg- 
ments 4  and  6.     How  far  from  the  center  is  P? 

Suggestion.    Use  Ex.  5. 

7.  In  two  similar  polygons  two  corresponding  sides  are  3  and  7. 
If  the  perimeter  of  the  first  polygon  is  45,  what  is  the  perimeter  of 
the  second? 

8.  The  perimeters  of  two  similar  polygons  are  32  and  84.  A  side 
of  the  first  is  11.  What  is  the  corresponding  side  of  the  second 
polygon  ? 


136 


PLANE  GEOMETRY. 


279.   Definitions.     In  a  right  triangle  ABC^  right-angled 


at  C,  the  ratio  —  is  called  the  sine  of  Z  A 
AB 

and  is  written  sin  A, 

If  any  other  point  b'  be  taken  on  the 

hypotenuse  or  the  hypotenuse  extended, 

and  a  perpendicular  b' c'  be  let  fall  to  AC^ 

&b'  _  CB 

AB'   ~  AB 


B'/ 


A^ 


c  c 


then 


(Why?) 


Likewise  in  A  ab"c",  in  which  AB'^  =  1  unit,  we  have 


&'b''      C'^B' 


=  c"b^'  =  sin  A. 


AB"  1 

Hence,  in  a  right  triangle  whose  hypotenuse  is  unity, 
the  length  of  the  side  opposite  an  acute  angle  is  the  sine  of  that 
angle. 

280.  Theorem.  The  ratio  of  the  sides  opposite  two 
acute  angles  of  a  triangle  is  equal  to  the  ratio  of  the 
sines  of  these  angles. 

Given  A  ABC  with  Z  A  and  Z  B  both  acute  angles. 

_,  ,1    ,  sin  A      a 

To  prove  that  ~ =  -. 

sin  B      0 

Proof:    Draw  the  perpendicular  j9. 

Then  sin  A  =\  and  sin  B  =^. 
0  ,       a 

Hence,  ^HL^=f^^  =  f    See  §241,  (3). 
sin  B      b      a      h  ^        '  ^  ^ 

Note.  The  definitions  of  §  279  and  the  theorem  of  §  280  are 
given  here  for  acute  angles  only.  In  trigonometry,  where  the  subject 
is  studied  in  full  detail,  they  are  extended  to  apply  to  any  angles 
whatever.  Other  ratios  called  cosines,  tangents,  etc.,  are  also 
introduced. 


MEASUREMENT  OF  LINE-SEGMENTS. 


137 


281.  The  theorem  of  §  280  is  of  great  importance  in 
finding  certain  parts  of  a  triangle  when  other  parts  are 
known.  By  careful  measurement  (and  in  other  ways) 
tables  may  be  constructed  giving  the  sine  of  any  angle. 


282. 


EXERCISES. 


1.  By  means  of  a  protractor  construct  angles  of  10°,  20°,  30°,  40°, 
50°,  60°,  70°,  80°,  and  measure  the  sine  of  each  angle,  and  so  construct 
a  table  of  these  sines. 

If  one  decimeter  is  used  as  a  unit  for  the  hypotenuse, 
then  the  length  of  the  side  opposite  Z  vl,  expressed  in 
terms  of  decimeters,  is  the  sine  of  the  angle  A. 

Notice  that  the  values  of  the  sines  are  the  same  no  matter  what 
unit  is  used,  but  in  general  the  larger  the  unit  the  more  accurately  is 
the  sine  determined. 

By  means  of  the  table  just  constructed  solve  the  follow- 
ing problems,  using  the  notation  of  the  figure : 

2.  Given  ZA=  30°,  B  =  80°,  b  =  12,  find  a,  c.  B 

Solution.    By  the  theorem,  §  280,  -  =  ^i^  • 

h      sin  B 

Substituting  the  values  h  =  12,  and  sin  A  =  sin  30°,  -4~        6=12       o 

sin  B  =  sin  80°  from  the  table,  we  find  a.     In  the  same  manner  find  c. 

3.  Given  a=l(y,ZA=  60°,  ZC  =  70°,  find  Z B,  h,  c. 

4.  A  lighthouse  L  is  observed  from  a  ship  S 
to  be  due  northeast.  After  sailing  north  9  miles 
to  S',  the  lighthouse  is  observed  to  be  35°  south 
of  east.  Find  the  distance  from  the  ship  to  the 
lighthouse  at  each  point  of  observation.  Use  §  284. 

5.  A  ladder  25  feet  long  rests  with  one  end  on 
the  ground  at  a  point  12  feet  from  a  wall.  At 
what  angle  does  the  ladder  meet  the  ground. 

6.  If  two  sides  and  the  included  angle  of  a 
triangle  are  known,  can  the  remaining  parts  be 
found  by  means  of  §  280? 


138  PLANE  GEOMETRY, 

283.  In  land  surveying  on  an  extensive  scale,  processes 
similar  to  that  used  on  the  preceding  page  are  constantly 
employed  in  finding  the  sides  of  triangles. 

To  begin  with,  a  level  piece  of  ground  is  selected  and  a 
line   AB   measured   with    great    care.    ^  e 

Then  a  point  C  is  selected,  and  /-ABC    \  ^"^-^^^c^-^"!    1>g 
and  Z.BAC  measured  very  accurately 
with   an   instrument.     Sides   AC  and 
BC  may  now  be   computed  by  means    -^  i> 

of  the  theorem,  §  280,  and  a  table  of  sines  (see  page 
opposite).  By  measuring  Z  BAC  and  Z  ACB,  CB  may  be 
computed.  By  this  process,  called  triangulating,  it  is 
possible  to  survey  over  a  large  territory  without  directly 
measuring  any  line  except  the  first. 

284.  The  saving  of  labor  afforded  by  this  indirect  method 
of  measuring  is  very  great,  and  especially  so  in  a  rough 
and  mountainous  country,  since  measuring  the  straight 
line  distance  from  one  mountain  peak  to  another  by  means 
of  a  measuring  chain  is  impossible. 

In  practice,  tables  of  logarithms  are  used  and  the  sines 
are  carried  out  to  a  larger  number  of  decimal  places,  but 
the  general  process  is  that  used  on  the  preceding  page. 

285.  EXERCISES. 

Using  the  table  on  the  next  page,  solve  the  following  examples: 

1.  Given  A  =  53%  B  =z  65%  a  =  11.5.     Find  b,  c,  and  Z  C. 

2.  Given  B  =  49°,  0  =  71°,  a  =  19.3.     Find  h,  c,  and  Z  A . 

3.  Given  A  =  65%  a  =  U,b=  12.     Find  ZB,  ZC,  and  c. 

Solution.     §H^  =  -^  or  sin  ^  =  sin  v4  x  -  =  .91  x  —  =  .78. 
Sin  A      a  a  14 

From  the  table  we  find  that  sin  51°  =.78.  Hence  £  =  5P.  ZC 
and  c  may  now  be  found  as  before. 


MEASUREMENT  OF  LINE-SEGMENTS. 


139 


4.    Given  A  =71°,  a  =  19.5,  b  =  17.     Find  ZB,  ZC,  and  c.     As  be- 
b 


fore,  sin  5  =  sin  yl  x 


sin  71°  X  -^  =  .95  X  — 
19.5  19.5 


.828.     From  the 


table  we  find  sin  5.5°  =  .82  and  sin  56°  =  .83.  But  sin  B  is  nearer  .83 
than  .82,  and  hence  B  =  56°  is  the  nearest  approximation  using  a 
degree  as  the  smallest  unit. 


Angle 

Sin 

Angle 

Sin 

Angle 

Sin 

Angle 

Sin 

Angle 

Sin 

Angle 

Sin 

0-^ 

0 

15° 

.26 

30° 

.50 

45° 

.71 

60° 

.87 

75° 

.97 

1° 

.02 

16° 

.28 

31° 

..52 

46° 

.72 

61° 

.87 

76° 

.97 

2° 

.03 

17° 

.29 

32° 

.53 

47° 

.73 

62° 

.88 

77° 

.97 

3° 

.05 

18° 

.31 

33° 

.54 

48° 

.74 

63° 

.89 

78° 

.98 

4° 

.07 

19° 

.33 

34" 

.56 

49° 

.75 

64° 

.90 

79° 

.98 

5° 

.09 

20° 

.34 

35° 

.57 

50° 

.77 

65° 

.91 

80° 

.98 

6° 

.10 

21° 

.36 

36° 

..59 

51° 

.78 

66° 

.91 

81° 

.99 

7° 

.12 

22° 

.37 

37° 

.60 

52° 

.79 

67° 

.92 

82° 

.99 

8° 

.14 

2.3° 

.39 

38° 

.62 

53° 

.80 

68° 

.93 

83° 

.99 

9° 

.16 

24° 

.41 

39° 

.63 

54° 

.81 

69° 

.93 

84° 

.99 

10° 

.17 

25° 

.42 

40° 

.64 

55° 

.82 

70° 

.94 

85° 

11° 

.19 

26° 

.44 

41° 

.66 

56° 

.83 

71° 

.95 

86° 

12" 

.21 

27° 

.45 

42° 

.67 

57° 

.84 

72° 

.95 

87° 

13° 

.22 

28° 

.47 

43° 

.68 

58° 

.85 

73° 

.96 

88° 

14° 

.24 

29° 

.48 

44° 

.69 

59° 

.86 

74° 

.96 

89° 

286.  Definition.  An  object  AB  is  said  to  subtend  an 
angle  APB  from  a  point  P  if  the  lines  PA  and  PB  are  sup- 
posed to  be  drawn  from  P  to  the  extremities  of  the  object. 


287. 


EXERCISES. 


1.  A  building  known  to  be  150  feet  high  is  seen  to  subtend  an  angle 
of  20°.  How  far  is  the  observer  from  the  building  if  he  is  standing 
on  a  level  with  its  base? 

2.  Show  how  to  find  the  distance  from  an  accessible  to  an  inacces- 
ible  point  by  means  of  §  280  and  the  table  of  sines. 

3.  A  flagstaff  is  125  feet  tall.  How  far  from  it  must  one  be  in 
order  that  the  flagstaff  shall  subtend  an  angle  of  25°? 


140  PLANE  GEOMETRY. 

PROBLEMS  OF  CONSTRUCTION. 

288.  Instruments.  In  addition  to  the  ruler,  compasses^ 
and  protractor  described  in  §  44,  the  parallel  ruler  is 
convenient  for  drawing  lines  through  given  points  parallel 
to  given  lines  without  each  time  making  the  construction 
of  §  101. 

Description.     R  and  R'  are  two  rulers  of  equal  length  and  width. 

AB  and  CD  are  arms  of  equal  length  pivoted  at       

the  points  A,  B,  C,  and  D,  making  AC  =  BD. 


Why   do    the    rulers,    when    thus    constructed,     i  ^g  "TPlig" 

remain  parallel  as  they  are  spread  ? 

289.  Segments  of  equal  length  can  be  laid  off  with 
great  accuracy  on  a  line-segment  by  means  of  the  com- 
passes. The  following  construction  makes  use  of  the 
compasses  and  parallel  ruler. 

290.  Problem.  To  divide  a  given  line-segment  into 
any  number  of  equal  parts. 


Construction.     Proceed   as   in   §  159,  Ex.  5,  using  the 
parallel  ruler  to  draw  the  parallel  lines. 
Give  the  construction  and  proof  in  full. 

Historical  Note.  The  idea  of  similarity  of  geometric  figures,  or 
"sameness  of  shape,"  is  one  of  early  origin,  as  is  also  the  simple 
theory  of  proportion.  It  was  probably  used  by  Pythagoras  to  prove 
the  famous  theorem  known  by  his  name.  (See  §  262.)  But  the  dis- 
covery by  him  of  the  incommensurable  case  (§  230)  showed  that  this 
theory  was  inadequate  for  the  rigorous  proof  of  all  theorems  on  simi- 
lar figures.  It  remained  for  Eudoxus,  the  teacher  of  Plato,  to  perfect 
a  rigorous  theory  of  ratio  and  proportion. 


MEASUREMENT  OF  LINE-SEGMENTS.  141 

Euclid,  following  his  predecessors,  deals  with  ratios  of  magnitudes 
ill  general  as  well  as  of  numbers.  Later  writers  have  frequently  in- 
sisted that  ratios  in  general  are  not  numbers.  But  nothing  is  gained 
by  this  procedure,  since  they  possess  all  the  properties  of  numbers.  In 
this  book  a  ratio  is  treated  simply  as  the  quotient  of  two  numbers. 
See  §§  238-242. 

291.  Problem.     Given  three  line-segments  m,  n,  p,  to 

construct  a  fourth  segment  q  such  that  —  =  - ;  that  is,  to 

n       q 

find  a  fourth  proportional  to  m,  n,  and  p. 

Construction.  Draw  two  indefi- 
nite straight  lines,  Ax  and  Ai/  mak- 
ing a  convenient  angle. 

On  Ai/  lay  off  AB  =  m,  BC  =  n. 
On  Ax  lay  off  AD=p.     Draw  BI).   ^~^d 
Through  C  draw  CE  II  BD.     Then  DE=  q  is  the  required 
segment. 

Give  the  proof  in  detail. 

292.  EXERCISES. 

1.  Apply  the  method  of  §  290  to  bisect  a  given  line-segment,  and 
compare  this  with  the  method  of  §  51. 

2.  Divide  a  line  7|  inches  long  into  11  equal  parts  by  the  method 
of  §  290,  and  compare  with  the  process  of  measuring  by  means  of  an 
ordinary  ruler  giving  inches  and  sixteenths. 

3.  Using  the  same  segments  as  in  §  291,  construct  a  segment  q 

such  that —=:- ;    also  such  that  -  =  — . 
p      q  p       q 

4.  Using  the  same  segments  as  in  the  preceding,  construct  a  seg- 
ment q  such  that  ^  =  - ;    also  such  that  ^  =  '" . 

m       q  n       q 

What  is  q  called  in  each  case  ? 

5.  If  the  given  segments  m,  n,  p  are  respectively  3|  inches,  ^f^  inches, 

4|  inches,  compute  q  such  that  —  =  ".     Also  construct  q  as  in  §  291 
and  compare  results.  ,        ^ 


142  PLANE  GEOMETRY. 

293.  Definitions.  If  three  numbers  «,  m,  and  b  are  such 
that  a :  m  =  m :  b,  then  m  is  the  mean  proportional  between 
a  and  5,  and  b  is  the  third  proportional  to  a  and  m. 

294.  Pkoblem.  To  construct  a  mean  proportional  be- 
tween two  given  segments  m  and  n. 


^y 


Construction.     On  an  indefinite  line  xy\2iyo^AB  =  m 
and  BC  =  n.     On  AC  as  a  diameter  construct  a  semicircle. 

At  B  erect  a  perpendicular  to  AC  meeting  the  semicircle 
in  D.     Then  BD  =  jt?  is  the  required  segment. 

Proof.     Draw  AB  and  CD  and  prove 

AABD  r^  ACBD.  ^ 

Complete  the  proof. 

295.  EXERCISES. 

1.  Show  that  in  §  277  DP  is  a  mean  proportional  between  PB 
and  PA. 

2.  If  in  the  above  problem  m  =  3,  and  n  =  5  show  that  the  con- 
struction gives  Vl5. 

3.  Show  how  to  construct  a  segment  p  =  VS,  also  p  =  V3,  p  =  V2, 
.by  means  of  the  above  process. 

4.  Show   how   to   construct  a  square   equal   in  area  to  a  given 
J  rectangle. 

<»  5.  Show  how  to  construct  on  a  given  base  a  rectangle  equal  in 
area  to  a  given  square. 

V  Suppose  AB  =  m  is  the  given  base  and  BD  =p  a.  side  of  the  given 
square,  the  two  segments  being  placed  at  right  angles  as  in  the  figure 
above.  The  problem  is  then  to  find  on  the  line  AB  the  center  of  a 
circle  which  passes  through  the  points  A  and  D.  To  do  this  connect  A 
and  D  and  construct  a  perpendicular  bisector  of  this  segment  meeting 
AC  in  a  point  which  is  the  center  of  the  required  circle.  BC  =  n  is 
then  the  required  side  of  the  rectangle  since  m-n  =  p^. 


MEASUREMENT  OF  LINE-SEGMENTS.  143 

296.    Problem.     To  divide  a  line-segment  into  three 
'parts  proportional  to  three  given  segments  m,  n,  p. 


A  D  E  B 

Construction.  Let  AB  be  the  given  segment.  Con- 
struct Ax  and  on  it  lay  off  m,  w,  p  as  shown  in  the  figure. 
Complete  the  figure  and  give  proof  in  full.    ^ 

297.  EXERCISES. 

1.  Divide  a  line-segment  5  inches  long  into  three  parts  proportional 
to  2,  3,  and  4. 

2.  Divide  a  segment  11  inches  long  into  parts  proportional  to  3,  5, 
7,  and  9. 

First  compute  the  lengths  of  the  required  segments,  then  construct 
them  and  measure  the  segments  obtained.  Compare  the  results. 
Which  method  is  more  convenient  ?     Which  is  more  accurate  ? 

3.  Divide  a  segment  9  inches  long  into  two  parts  proportional  to  1 
and  V2. 

Also  compute  the  required  segments.  Which  is  more  convenient? 
More  accurate? 

4.  Divide  a  segment  whose  length  is  vTl  into  two  parts  propor- 
tional to  V'2  and  V5. 

First  construct  the  segments  whose  lengths  are  \/2,  a/5,  and  Vll. 
Also  compute  the  required  segments. 

5.  Divide  a  given  line-segment  into  parts  proportional  to  two  given 
segments  m  and  n,  (a)  if  the  division  point  falls  on  the  segment;  (ft) 
if  the  division  point  falls  on  the  segment  produced.     See  §  252. 

'  6.  A  triangle  is  inscribed  in  another  by  joining  the  middle  points 
of  the  sides. 

(a)  What  is  the  ratio  of  the  perimeters  of  the  original  and  the 
inscribed  triangles? 

(6)  Is  the  inscribed  triangle  similar  to  the  original? 


144  PLANE  G:E0METRV. 

298.    Problem.     On  a  given  line-segment  as  a  side 
construct  a  triangle  similar  to  a  given  triangle. 


f 


Construction.  Let  ABC  be  the  given  triangle  and  a'b' 
the  given  segment,  and  let  it  correspond  to  AB  of  the 
given  triangle. 

Construct  Z  2  =  Z  1  and  Z  4  =  Z  3  and  produce  the  sides 
till  they  meet  in  &,  Then  A'b'c'  is  the  required  tri- 
angle (Why?). 

299.  EXERCISES. 

1.  In  the  problem  of  §  298  construct  on  A'B'  a  triangle  similar  to 
ABC  such  that  A'B'  and  BC  are  corresponding  sides. 

2.  Show  that  on  a  given  segment  three  different  triangles  may  be 
constructed  similar  to  a  given  triangle. 

3.  Solve  the  problem  of  §  298  by  making  Z  2  =  Z  1  and  construct- 

AB       A  C 

ing  A'C  so  that  — —  =  -777-7 *    Give  the  solution  and  proof  in  full. 
A  B       A  Ly 

4.  On  a  given  segment  as  a  side  construct  a  polygon  similar  to  a 
given  polygon,  by  first  dividing  the  given  polygon  into  triangles  and 
then  constructing  triangles  in  order  similar  to  these.     Apply  §  270. 

5.  "On  a  given  segment  construct  a  polygon  similar  to  a  given 
polygon  in  a  manner  analogous  to  the  method  used  in  Ex.  3  for  con- 
structing a  triangle  similar  to  a  given  triangle. 

6.  ABC  is  a  right  triangle  with  the  hypotenuse  AB,  and 
CD1.AB.  Prove  AC  2b  mean  proportional  between  AB  and  AD 
and  likewise  CB  a  mean  proportional  between  AB  and  DB. 

7.  Use  the  theorem  of  §  277  to  construct  a  square  equal  in  area  to 
that  of  a  given  rectangle. 

8.  The  tangents  to  two  intersecting  circles  from  any  point  in  their 
common  chord  produced  are  equal.     Use  §  277. 


MEASUREMENT  OF  LINE-SEGMENTS. 

J 

9.  Ill  the  figure  AD  and  BE  are  tangents  at  the 
extremities  of  a  diameter.  If  BD  and  AE  meet  in  a 
point  C  on  the  circle  prove  that  AB  is  a  mean  pro- 
portional between  AD  and  BE. 

10.  The  greatest  distance  to  a  chord  8  inches  long 
from  a  point  on  its  intercepted  arc  (minor  arc)  is  2 
inches.     Find  the  diameter  of  the  circle.     Use  §  272. 

11.  At  the  extremities  of  a  chord  AB  tangents 
are  drawn.  From  a  point  in  the  arc  AB  perpen- 
diculars PC,  PD,  PF  are  drawn  to  the  chord  and 
the  tangents.  Prove  that  PC  is  a  mean  propor- 
tional between  PD  and  PF. 

Suggestion.'  Draw  segments  AP  and  BP  and 
prove  AAPF^ABCP  and  A  .1  PC  ~  BPD.     Then 

^.^and^  =  ^. 
PB     PC  PB     PD 


145 


SUMMARY  OF  CHAPTER  III. 

1.  Make  a  list  of  the  definitions  in  Chapter  III. 

2.  Make  a  list  of  the  theorems  on  proportional  segments  involving 
triangles. 

3.  State  the  various  conditions  which  make  triangles  similar. 

4.  State  the  conclusions  which  can  be  drawn  when  it  is  known 
that  two  triangles  are  similar. 

5.  State  the  theorems  on  proportional  segments  involving  poly- 
gons. 

6.  State  the  theorems  on  proportional  segments  involving  straight 
lines  and  circles. 

7.  State  ill  the  form  of  theorems  the  various  ways  in  which  a 
proportion  may  be  taken  so  as  to  leave  the  four  terms  still  in 
proportion. 

8.  Under  what  conditions  is  a  segment  a  mean  proportional  be- 
tween two  given  segments.     For  instance,  see  §  277. 

9.  Make  a  list  of  the  problems  of  construction  in  Chapter  III. 
10.    State  some  important  applications  of  the  theorems  on  propor- 
tion.    (Return  to  this  question  after  studying  those  which  follow.) 


146 


PLANE  GEOMETRY, 


PROBLEMS  AND  APPLICATIONS. 

1.  The  middle  points  of  adjacent  sides  of  a 
square  are  joined. 

(a)  Prove  that  the  inscribed  figure  is  a  square. 
(&)    What  is  the  ratio  of  the  inscribed  and  the 
original  squares. 

(c)  If  a  side  of  the  original  square  is  a,  find  a 
side  of  the  inscribed  square. 

(d)  If  a  side  of  the  inscribed  square  is  ft,  find  a 
side  of  the  original  square. 

2.  Two  strips  intersect  at  right  angles. 

(a)  If  the  width  of  each  strip  is  3  inches,  what 
is  the  largest  square  which  can  be  placed  on  them 
so  that  its  sides  will  pass  through  the  corner 
points  as  shown  in  the  figure?  (The  corners 
will  bisect  the  sides  of  the  square.) 

(ft)  What  is  the  side  of  this  square  if  the 
width  of  each  strip  is  a  ? 

(c)  Find  the  side  of  the  square  if  the  width 
of  one  strip  is  3  and  that  of  the  other  5. 

(of)  Find  the  side  of  the  square  if  the  width 
of  one  strip  is  a  and  that  of  the  other  ft. 

3.  In  the  figure  ABCD  is  a  square  and 
EFGH  •••  is  a  regular  octagon. 

(a)  Show  that  A  E,  EF,  FB  are  proportional 
to  1,  V2,  1.     (Assume  ^jEJ  =  1  an^  find  NE.) 

^    (ft)  Ifyl^^a,showthat  — =  ^ +^^  =  — • 

(c)  Find  Z  FEN.     See  §  163,  Ex.  2. 
|,  (d)  Show  that   AO  =  AF,  and  hence  that 
the    regular    octagon    may   be   constructed   as 
shown  in  the  figure. 

(e)  Show  that  ZAOE  =  22^°  by  §  219,  and 
use  this  to  make  another  proof  that  EFGH  ... 
is  a  regular  octagon. 

4.  Find  the  area  of  a  square  whose  diagonal 
is  a. 


Tile  Pattern. 


17T 


Tf 


Tile  Pattern. 


MEASUREMENT   OF  LINE-SEGMENTS. 


147 


5.  If  a  side  of  a  regular  octagon  is  a,  find  its  area. 

6.  A  floor  is  tiled  with  regular  white 
octagons  and  black  squares  as  shown  in 
the  design.  What  per  cent  (approxi- 
mately) of  the  floor  is  black? 

7.  Divide  a  circle  into  eight  equal  parts 
and  join  alternate  division  points. 

(a)  Prove  that  LMNPQ  •••  is  a  regular 
octagon. 


Cut  Glass  Design. 

(?>)  If  the  diameter  of  the  circle  is  a,  show  that  LE  —  ~(Vi 
Suggestion.     Find  HE  and  then  use  Ex.  3,  (a). 

"^  8.  Problem.  Inscribe  a  square  in  a  given  semicircle. 
Construction.  Let  AB  be  the  diameter  of  the 
given  semicircle.  At  B  construct  a  perpendicular  to 
AB,  making  BD==AB.  Connect  D  with  the  center  (7, 
meeting  tlie  circle  at  E.  Let  fall  EF  perpendicular 
to  AB.  Then  EF  is  a  side  of  the  required  square. 
Complete  the  figure  and  make  the  proof  by  showing 
that  EF  =  2  CF. 

9.    Problem.     Inscribe  a  square  in  a  given  triangle. 

Construction.  Let  ABC  be  the  given  tri- 
angle. Draw  CD  parallel  to  AB,  making  CD  = 
CO.     Complete  the  square  CDEO. 

Draw  DA  meeting  CB  in  F.  Draw 
FG  II  AB,  GH  li  CO,  and  FK  II  CO.  Then 
FGHKr^  CDEO  and  hence  is  the  required  square. 


1). 


-^E 


148 


PLANE  GEOMETRY. 


t 


f 


Touch  down  ®°'^^  P°«ts 


^  ,  10.    The  sides  of  a  triangle  are  13,  17,  19.     Find  the  lengtlis  of  the 
segments  into  which  the  angle  bisectors  divide  the  opposite  sides. 

11.  The  angles  of  a  triangle  are  30°,  (J0°,  90°.  Find  the  lengths  of 
the  segments  into  which  the  angle  bisectors  divide  the  opposite  sides 
if  the  hypotenuse  is  10. 

12.  Prove  that  the  perpendicular  bisectors  of  all  the  sides  of  a 
polygon  inscribed  in  a  circle  meet  in  a  point. 

13.  An  equilateral  polygon  inscribed  in  a 
circle  is  equiangular. 

.  14.    The  goal  posts  on  a  football  field  are  18| 

feet  apart.  If  in  making  a  touchdown  the  ball 
crosses  the  goal  line  25  ft.  from  the  nearest 
goal  post,  how  far  back  should  it  be  carried  so 
that  the  goal  posts  shall  subtend  the  greatest 
possible  angle  from  the  place  where  the  ball  is 
'placed?    See  Ex.  22,  page  111,  and  §  277. 

F 

15.  Solve  the  preceding  problem  if  the  touchdown  is  made  a  feet 

from  the  nearest  goal  post,  and  thus  obtain  a  formula  by  means  of 
which  the  distance  may  be  computed  in  any  case. 

16.  A  triangle  has  a  fixed  base  and  a  constant 
vertex  angle.  Show  that  the  locus  of  the  vertex  is 
two  arcs  whose  end-points  are  the  extremities  of  the 
base.     See  Ex.  20,  page  111. 

Note  that  the  locus  also  includes  an  arc  on  the 
opposite  side  of  AB  from  the  one  shown  in  the 
figure. 

It  follows  that  if  two  points  A  and  B  subtend  the  same  angle  from 
P  and  from  Cf,  then  a  circle  may  be  passed  through 
A,B,P,Q. 
»|       17.    In  the  figure,  A,  B,    F    lie  in    the   same 
^traight  line,  as  do  also  D,  C,  F,  and  Z  1  =  Z2. 

(a)  If  BF  =8,CF=  9,  and  DC  =  12,  find  AB. 

(b)  JiAB  =  100,  AF  =  250,  CF  =  60,  find  DC. 
Note.     By  holding  a  small  object,  say  a  pencil, 

at  arm's  length,  and  sighting  across  the  ends  of 
it,  we  may  determine  approximately  whether  two 
given  objects  subtend  the  same  angle. 


MEASUREMENT  OF  LINE-SEGMENTS. 


149 


f  18.  Not  having  any  instruments,  an  engineer  proceeds  as  follows 
to  obtain  approximately  the  distance  from  an  accessible  point  B  to  an 
inaccessible  point  A .     Walking  from  B  along  ^ 

the  line  AB  he  takes  50  steps  to  F.  Then  he 
walks  in  a  convenient  direction  50  steps  to  C, 
and  notes  that  A  and  B  subtend  a  certain  angle. 
He  then  proceeds  along  the  same  straight 
line  until  he  reaches  a  point  D  at  which  A 
and  B  again  subtend  the  same  angle  as  at  C. 
He  then  concludes  that  DC  =  AB.  Is  th;s 
conclusion  correct?     Give  proof. 

^  19.  If  the  height  of  a  building  is  known,  show  how  the  method  of 
Ex.  18  can  be  used  to  determine  the  height  of  a  flagstaff  on  it. 
*  20.  A  building  is  130  feet  high,  and  a  flagstaff  on  the  top  of  it  is 
60  ft.  high;  130  feet  from  the  base  of  the  building  in  a  horizontal 
plane,  the  flagstaff  subtends  a  certain  angle.  How  far  from  the  build- 
ing along  the  same  line  is  there  another  point  at  which  the  staff  sub- 
tends the  same  angle?  At  what  distance  does  it  subtend  the  same 
angle  as  it  does  at  300  feet? 

^       21.   If  the  diagonals  of  an  isosceles  trapezoid  are  drawn,  what  similar 

triangles  are  produced? 
s'       22.    Find  the  locus  of  points  at  a  fixed  distance  from  a  given  tri- 
angle, ^ways  measuring  to  the  nearest  point  on  it. 
23.   The  line  bisecting  the  bases  of  any  trapezoid 
passes   through    the   point   of   intersection   of   its 
diagonals. 

Suggestion.     Let  E  bisect  DC,  and  draw  EO 
meeting  AB  in  F.     Prove  AF=  FB. 

I     24.   If  two  triangles  have  equal  bases  on  one  of  two  parallel  lines, 
and  their  vertices  on  the  other,  then  the  sides  of  these  triangles  intercept 
equal  segments  on  a  line  parallel  to  these  and  lying  between  them. 
i      25.    A  segment   bisecting  the  two  bases  of  a         D       P       c 
'  trapezoid  bisects  every  segment  joining  its  othev 
two  sides  and  parallel  to  the  bases. 

(Prove  EF^BG  and  FK^  KH,  using  Ex.  24.)     *^  ""' 

26.   In  a  triangle  lines  are  drawn  parallel  to  one  side,  forming  trap- 
ezoids.    Find  the  locus  of  the  intersection  points  of  their  diagonals. 


E^ 


150 


PLANE  GEOMETRY. 


27.  In  the  figure  D,  E,  F  are  the  middle  points  of  the  sides  of  an 
equilateral  A  ABC.  Arcs  are  constructed  with  centers  A,  B,  C  as 
shown  in  the  figure. 

(a)  Prove  that  these  arcs  are  tangent  in  pairs  at 
the  points  D,  E,  F. 

(h)  Construct  a  circle  tangent  to  the  three  arcs, 
(c)  What  axes  of  symmetry  has  this  figure? 

28.  In  the  figure  ABC  is  an  equilateral  triangle. 
Arcs  AB,  BC,  and  CA  are  constructed  with  C,  A, 
and  B  as  centers.     A  F,  BE,  CD  are  altitudes  of 
the  triangle. 

Prove  QOG  tangent  to  AB,  BC,  and  CA. 

29.  In  the  equilateral  triangle  ABC,  AB,  BC, 
and  CA  are  constructed  as  in  the  preceding  ex- 
ample. D^,  EF,  FD  are  constructed  as  in 
Ex.  27.  The  figure  is  completed  making  ADE, 
DFB,  and  EEC  similar  to  ABC. 

(a)  Construct  a  circle  tangent  to  a1),  DE,  and 
EA  as  shown  in  the  figure. 

(b)  Construct  a  circle  tangent  to  ^D,  DF, 
and  FE. 

•^  30.  In  the  figure  A  CB  is  a  semicircle.  Arcs 
DE  and  DF  are  constructed  with  B  and  A  as 
centers,     li  AD  =  4:  feet,  find  the  radius  of  O  OC. 

Solution.     Show  that  the  value  of  r.  is  derived 
from  the  equation  (4  +  r)^  =  (4  -  r)2  +  4^.  ^         ^ 

31.  Construct  the  accompanying  design.  Notice  that  the  points 
A ,  B,  M,  N,  F,  E  divide  the  circle  into  six  equal  arcs.     See  Ex.  5, 


From  Boynton  Cathedral,  England. 


MEASUREMENT  OF  LINE-SEGMENTS. 


151 


The 


§  163.  The  arcs  A  0,  OB,  etc.,  and  the  circle  with  center  0',  are 
constructed  as  in  Ex.  28.  The  small  circle  with  center  at  Q  is  con- 
structed as  in  the  preceding  example. 

►.       32.   ^5 C  is  an  equilateral  arch  and  AEB  is  a  semicircle. 

(a)  If  AB  (the  span  of  the  arch)  is  10  feet,  C 
find   the  radius  of   the   small  circle   tangent   to 
the  semicircle  and  the  arcs  of  the  arch. 

(b)  Find  the  radius  OE  of  the  circle  ii  AB  =  s. 

(c)  If  OE  =  2,  find  AB. 

5      33.    The  accompanying  design  consists  of  three       A 
semicircles  and  three  circles  related  as  shown  in  the  figure. 

(a)  If    AB  =  12,   find    OQ  by   using  the    triangle    OSD. 
method  is  similar  to  that  used  in  Ex.  30. 

(b)  Using  the  right  triangle  DRO',  find 
O'x.  Notice  that  in  order  to  have  the  circles 
with  the  centers  0  and  0'  tangent  to  each 
other  the  sum  of  their  radii  must  be  equal 
to  RD. 

(c)  If    AB  =  s    show     that    0(2=-.    and 

O'x  =  ~,  and  construct  the  figure. 

34.  Upon  a  given  segment  AB  construct 
the  design  shown  in  the  figure.  Notice  that 
it  consists  of  the  two  preceding  figures  put 
together.  Compare  the  radii  of  the  circles 
in  this  design. 


A        E         H  B 

Ospedale  Maggiore,  Milau. 
•     35.    Between  two  parallel  lines  construct 
semicircles  and  circles  as  shown  in  the  figure. 

Solution.     Suppose   the   construction   made.      Let  HE  =  r,  and 


OQ  =  r'.     Using  the  triangle  EHO,  show  that 
struct  the  fig'ure. 


and  then  con- 


152 


PLANE  GEOMETRY. 


\        36.   Between  two  parallel  lines  construct  circles  and  equilateral 
arches  as  shown  in  the  figure. 


The  Doge's  Palace,  V'euice. 
Hint.  AQ  =  2  PR  and  AH  =  3  PH. 

Hence  HQ^  =  WPH^  -  YPH^  =  TPH  \ 

That  is,  PH  :HQ  =  1'.  V5.  (Why  ?) 

Now  divide  AC  into  two  segments  proportional  to  1  and  Vo,  and 
so  construct  the  figure. 

C 


37.   ABC  is   an   equilateral  arch.     AD  =  DB,  AF=EB=^AB.  ^ 
Semicircles  are  constructed  with  E  and  F  as  centers,  and  radius  FB.    • 

(a)  If  .4J5  =  s,  find  OH  and  construct  the  figure.  j 

Solution.     Let   OH 
F0  =  -  +  r. 


r.      Then   BO  =  s  -  r,  FB='-,  DF  =  -, 

3  6     - 


i 


From  A  ODB,    Ojf  =  Olf  -  1)B^  =  (.s  -  r)' 


From  A  ODF,    Olf  =  OF^  -  DF^  =  [t  +  r 

From  (1)  and  (2)    (.  _  r)^  _  (0' =  (|  + r)'- (01 

Hence,  show  that  r  —  \s. 

Construct  the  figure. 

(6)  If  in  the  preceding  0H=^  feet,  find  AB  and  AF. 


(1), 
(2j 


MEASUREMENT  OF  LINE-SEGMENTS. 


153 


7~^ 

IR. 

/ 


m_M: 

AT 


saL 


WQ 


38.  Between  two  parallel  lines  construct 
circles  and  semicircles  having  equal  radii  as 
shown  in  the  figure. 

Prove  that  the  ratio  —  =  — ,  RLK  being 
HQ      RS  ^ 

an  equilateral  triangle  and  KS  =  SL. 

Divide  CA  in  the  ratio  of  SL  :  RS^ 

39.  Through  a  fixed  point  on  a  circle  chords  are  drawn  and  each 
extended  to  twice  its  length.  Find  the  locus  of  the  end-points  of 
these  segments.     Compare  Ex.  6,  §  214. 

40.  If  a  quadrilateral  is  circumscribed  about  a  circle,  show  that 
the  sums  of  its  pairs  of  opposite  sides  are  equal. 

41.  On  a  diameter  produced  of  a  given  circle,  find  a  point  from 
which  the  tangents  to  the  circle  are  of  a  given  length.  Solve  this 
problem  by  construction,  and  also  algebraically. 

42.  Compare  the  perimeters  of  equilateral  triangles  circumscribed 
about  and  inscribed  in  the  same  circle. 

43.  In  a  given  square  construct  semicircles  each  tangent  to  two 
sides  of  the  square  and  terminating  on  the  diameters  of  the  square. 

Construction.  Connect  E  and  F  two  ex- 
tremities ol  diameters  and  on  EF  as  a  diame- 
ter construct  a  semicircle  with  center  at  0'. 
Draw  O'H  perpendicular  to  AB  meeting  the 
arc  in  H.  Draw  OH  meeting  AB  in  K.  Draw 
KO"  perpendicular  to  AB  meeting  the  diagonal 
^C  in  0".  Then  0"  is  the  center  of  the  re- 
quired circle. 


Proof.     Draw  0"L\\  FE.     We  need  to  prove  that  0"K  =0"L. 

00"  _  0"L 
00'  ~  O'E 
00"  ^0"K 
00'  ~  O'H 


A  00' E  ~  A  00"L,  and  hence 
Also    A  00' H  ~  A  00"K,  and  hence 


(Why?)    (1) 
(2) 


V  ,^^        A  /ON  0"L       0"K 

From  (1)  and  (2)  ^^^  =  ^  • 

But  O'E  =  O'H,  and  hence  0"  L  =  0"K. 

This  figure  occurs  in  designs  for  steel  ceilings. 


(Why?) 


CHAPTER   IV. 

AREAS  OP  POLYGONS. 

AREAS  OF  RECTANGLES. 

300.  Heretofore  certain  properties  of  plane  figures  have 
been  studied,  such  as  congruence  and  similarity^  but  no 
attempt  has  been  made  to  measure  the  extent  of  surface 
inclosed  by  such  figures.  For  this  purpose  we  first  con- 
sider the  rectangle. 

301.  The  surface  inclosed  by  a  rectangle  is  said  to  be 
exactly  measured  when  we  find  how  many  times  some 
unit  square  is  contained  in  it. 

E.g.  if    the  base   of  a  rectangle  is  five    units      ^ 
long  and  its  altitude  three  units,  its  surface  con- 
tains a  square  one  unit  on  a  side  fifteen  times. 

302.  The  number  of  times  which  a  unit  square  is  con- 
tained in  the  surface  of  a  rectangle  is  called  the  numerical 
measure  of  the  surface,  or  its  area. 

We  distinguish  three  cases. 

303.  Case  1.  If  the  sides  of  the  given  rectangle  are  in- 
tegral multiples  of  the  sides  of  the  unit  square,  then  the 
area  of  the  rectangle  is  determined  by  finding  into  how 
many  unit  squares  it  can  be  divided. 

Thus  if  the  sides  of  a  rectangle  are  m  and  n  units  respectively,  then 
it  can  be  divided  into  m  rows  of  unit  squares,  each  row  containing  n 
squares.     Hence  the  area  of  such  a  rectangle  is  m  x  n  unit  squares,  that   i 
is,  in  this  case,  ^^^^  ^  ^^^^  ^  ^j^.^^^^  ^^  ^    1 

154 


--t---i— |— -|--- 
— i — I — 1.__< — 


AREAS   OF  POLYGONS. 


155 


304.  Case  2.  If  the  sides  of  the  rectangle  are  not  inte- 
gral multiples  of  the  side  of  the  chosen  unit  square,  but 
if  the  side  of  this  square  can  be  divided  into  equal  parts 
such  that  the  sides  of  the  rectangle  are  integral  multiples 
of  one  of  these  parts,  then  the  area  of  the  rectangle  may 
be  expressed  integrally  in  terms  of  this  smaller  unit  square, 
a,nd  fractionally  in  terms  of  the  original  unit. 

For  example,  if  the  base  is  3.4  deci- 
meters and  the  altitude  2.6  decimeters, 
then  the  rectangle  cannot  be  exactly  di- 
vided into  square  decimeters,  but  it  can 
be  exactly  divided  into  square  centi- 
meters. Each  row  contains  34  centi- 
meters and  there  are  26  such  rows. 

Hence,  the  area  is  34  x  26  =  884 
small  squares  or  8.84  square  decimeters. 

But  3.4  X  2.6  =  8.84.     Hence,  in  this  case  also, 

area  =  base  x  altitude.  (2) 

305.  Case  3.  If  a  rectangle  is  such  that  there  exists  no 
common  measure  whatever  of  its  base  and  altitude,  then 
there  is  no  surface  unit  in  terms  of  which  its  area  can  be 
exactly  expressed.  But  by  choosing  a  unit  sufficiently 
small  we  may  determine  the  area  of  a  rectangle  which 
differs  as  little  as  we  please  from  the  given  rectangle. 

E.g.  if  the  base  is  5  inches  and  the  altitude  is  v^5  inches,  then  the 
rectangle  cannot  be  exactly  divided  into  equal  squares,  however  small. 

But  since  V5  =  2.2361  •••,  if  we  take  as  a  unit  of  area  a  square 
whose  side  is  one  one-thousandth  of  an  inch,  then  the  rectangle  whose 
base  is  5  inches  and  whose  altitude  is  2.236  inches  can  be  exactly  meas- 
ured as  in  cases  1  and  2,  and  its  area  is  5  x  2.236=  11.18  square  inches. 

The  small  strip  by  which  this  rectangle  differs  from  the  given  rec- 
tangle is  less  than  .0002  of  an  inch  in  width,  and  its  area  is  less  than 
5  X  .0002  =  .001  of  a  square  inch. 

By  expressing  V5  to  further  places  of  decimals  and  thus  using 
smaller  units  of  area,  successive  rectangles  may  be  found  which  differ 
less  and  less  from  the  given  rectangle. 


156  PLANE  GEOMETRY. 

306.  An  area  thus  obtained  is  called  an  approximate 
area  of  the  rectangle. 

For  practical  purposes  the  surface  of  a  rectangle  is 
measured  as  soon  as  the  width  of  the  remaining  strip  is 
less  than  the  width  of  the  smallest  unit  square  available. 

From  the  foregoing  considerations  we  are  led  to  the 
following  preliminary  theorem ; 

307.  Theorem.  The  area  of  a  rectangle  is  equal  to 
the  product  of  its  base  and  altitude. 

308.  The  argument  used  above  shows  that  the  theorem 
(§  307)  holds  for  all  rectangles  used  in  the  process  of 
approximation,  and  hence  it  applies  to  all  practical  meas- 
urements of  the  areas  of  rectangles. 

AREAS  OF   POLYGONS. 

309.  From  the  formula  for  rectangles, 

area  =  base  x  altitude, 

we  deduce  the  areas  of  other  rectilinear  figures  by  means 
of  the  principle  : 

Two  rectilinear  figures  are  equivalent  (that  is,  have  the 
same  area)  if  they  are  congruent,  or  if  they  can  be  divided 
into  parts  which  are  congruent  in  pairs. 

E.g.  the  two  figures  here  shown  are  equiv-  Aiv  / 

alent  since  AT  and  III  are  congruent  respec-         y'^\v^y^     /n'/ 
tively  to  II  and  IV.  ^ /-— -i/ 

There  are  figures  of  equal  area  which  cannot  be  thus  divided. 
Thus  a  rectangle  with  base  2  and  altitude  1  is  equal  in  area  to  a 
square  whose  side  is  \/2,  though  these  figures  cannot  be  divided  in  the 
manner  stated  above.  However,  the  test  specified  in  the  principle  is 
sufficient  for  our  present  purposes. 

The  symbol  =  joining  two  polygons  means  equivalent 
or  equal  in  area. 


ABEAS   OF  POLYGONS.  157 

310.    Theorem.     The  area  of  a  parallelogram  is  equal 
to  the  product  of  its  base  and  altitude. 


III 


Given  the  parallelogram  ABCD  whose  base  is  AB  and  whose 
altitude  is  AE. 

To  prove  that  area  ABCD  =  AB  x  AE. 

Proof :  Draw  BF  ±  to  DC  produced,  forming  the  rec- 
tangle ABFE,  whose  base  is  AB  and  altitude  AE. 

Then  area  ABFE  =  AB  x  AE  (Why  ?). 

If  now  we  prove  Al^AlI,  then  the  parallelogram  is 
composed  of  parts  I  and  III  which  are  congruent  respec- 
tively to  parts  II  and  III  of  the  rectangle,  and  hence  the 
parallelogram  and  rectangle  are  equal  in  area.  Give  this 
proof  in  full. 

311.  EXERCISES. 

1.  In  the  figure  CF  is  perpendicular  to  AB  and  AE  to  CB. 
Prove  that  ABxCF  =  BCxAE.  ^^  ^, 

Suggestion.     (Show  that  A^5J5;  ~  AC^F.) 

It  follows  that  the  same  result  is  obtained  if  either 
of  two  adjacent  sides  of  a  parallelogram  is  taken 


as  the  base.  ^  ^  ^ 

2.  Construct  the  figure  described  in  Ex.  1  in  such  manner  that 
the  point  E  falls  on  BC  extended  and  prove  the  theorem  in  that  case. 

In  the  following  \  and  l^  are  parallel  lines. 

3.  Two  parallelograms  have  equal  bases  lying  on  l^  and  l^.     Show 
that  they  are  equivalent. 

4.  One  base  of  a  parallelogram  is  fixed  on  l^  and  the  other  moves 
along  I.,.     Does  the  area  change? 


158 


PLANE  GEOMETRY. 


312.    Theorem.     The  area  of  a  triangle  is  equal  to 
one  half  the  product  of  its  base  and  altitude. 

Given  the  A  ABC  whose  altitude 
upon  the  side  AB  is  CD. 

To  prove  that  the  area  of 
AABC=  I  (^AB  X  CD). 

Proof  :  Let  G  be  the  middle 
point  of  AC.  Complete  the 
parallelogram  ABFG. 

Prove  that  A  CGH  ^  A  BFH^  and  hence  show  that  A  ABC 
and  O  ABFG  have  equal  areas. 

Hence  the  area  of  AABC=  ED  x  AB. 

Hence  the  required  area  is  J  (^ii  x  CD). 


But  ^D  =  I  CD. 


313. 


EXERCISES. 


1.  Prove  the  theorem  of  §  312  us- 
ing the  accompanying  figure. 

2.  Prove  this  theorem  also  by  draw- 
ing a  figure  in  which  the  given  triangle 
is  one  half  of  a  parallelogram. 

3.  If    the    middle   points    of    two    ^i  B 
adjacent  sides  of  a  parallelogram  are  joined,  a  triangle  is  formed 
whose  area  is  equal  to  one  eighth  of  the  area  of  the  parallelogram. 

In  the  following  exercises  l^  and  l^  are  parallel  lines. 

4.  Show  that  two  triangles  are  equivalent  if  their  vertices  lie  in 
^2  and  their  equal  bases  in  ly 

5.  A  triangle  has  a  fixed  base  in  L^.  If  its  vertex  moves  along  Zj, 
what  can  you  say  of  its  area? 

€.  If  the  base  of  a  triangle  is  fixed  and  if  its  vertex  moves  so  as  to 
preserve  the  area  constant,  what  is  the  locus  of  the  vertex  ? 

7.  A  line  is  drawn  from  a  vertex  of  a  triangle  to  the  middle  point 
of  the  opposite  side.    Compare  the  areas  of  the  triangles  thus  formed. 

If  a  segment  is  drawn  from  the  vertex  to  the  point  P  in  the  base 
AB,  show  that  the  areas  of  the  triangles  are  in  the  ratio  AP  :  PB. 


AREAS   OF  POLYGONS. 


159 


314.  Theorem.  The  area  of  a  trapezoid  is  equal  to 
one  half  the  sum  of  its  bases  multiplied  by  its  altitude. 

Proof  :  Draw  a  diagonal,  thus 
forming  two  triangles  having  a 
common  altitude.  Form  the  ex- 
pression for  the  sum  of  the  areas 
of  these  triangles.  ^  ^ 

Give  the  proof  in  detail. 

315.  Theorem.  The  square  erected  on  the  sum  of  two 
line-segments  as  a  side  is  equal  to  the  sum  of  the  squares 
erected  on  the  two  segments  separately,  plus  twice  the  rec- 
tangle whose  base  is  one  segment  and  whose  altitude  is  the 
other  segment. 

Proof  :  Let  a  and  b  represent  the  numerical 
measures  of  the  two  line-segments.  Then,  the 
square  erected  upon  the  segment  a -{-  h  may  be 
subdivided  as  shown  in  the  figure,  giving 

(a  -\-by  =  a'^  +  2ab  +  b\ 
Give  the  construction  and  proof  in  full. 


316.  EXERCISES. 

1.  Show  by  a  figure  that 

(a-\-b  +  c)2  =  a2  +  5M-  c2  +  2  a6  +  2  ac  +  2  be. 

2.  If  ABC  is  any  triangle  and  AD,  BE,  and  CF  are  its  altitudes, 


show  that  i^  =  4£,  and  ^ 
CF 


AB 
AC 


BE      BC 

Hence  show  that 

AD  X  CB  =  BE  y.  CA  =  CF  x  AB. 

3.  The  side  AB  oi  CD  A  BCD  is  fixed.  What 
is  the  locus  of  the  points  C  and  D  if  CD  moves  so  as 
to  leave  the  area  of  the  parallelogram  fixed  ? 


160 


PLANE  GEOMETRY. 


317.  Theorem.  The  square  erected  on  the  difference 
of  two  line-segments  as  a  side  is  equal  to  the  sum  of  the 
squares  erected  on  the  two  segments 
separately,  minus  twice  the  rectangle 
whose  base  is  one  of  the  segments  and 
whose  altitude  is  the  other. 

Proof :    Use  the  figure  to  show  that 
(a  _  6)2  =  a2  +  52  _  2  «5. 

Give  the  construction  and  proof  in  detail. 


a 

.  f.-' 

ab        y 

b 

a-b       \ 
a-b      I  b 

318. 


EXERCISES. 


ah 


•*     1.   Show  by  a  figure  that 
*-  (a  —  b)(a  —  c)  =  a^  +  be 

2.  Likewise,  show  that  (a  -\-  b)(a  —  b)  =  a^  —  b^. 

3.  Show  by  counting  squares  ih  the  accompany- 
ing figure  that  in  case  of  an  isosceles  right  triangle 
the  square  constructed  on  the  hypotenuse  is  equal 

-4n  area  to  the  sum  of  the  squares  on  the  two  legs. 

4.  In  case  the  triangle  is  scalene,  show 
by  counting  squares  and  congruent  parts 
of  squares  that  the  square  constructed  on 
the  hypotenuse  is  equal  in  area  to  the  sum 
of  the  squares  on  the  other  two  sides. 

5.  Using  the  third  figure,  show  that  the 
same  theorem  holds  for  any  right  triangle. 

Hint.  Call  the  legs  of  the  given  triangle 
a  and  b.  Describe  the  construction  of  the 
auxiliary  lines  and  give  the  proof  in  full. 

This  theorem  is  proved  again  in  the  next 
paragraph  and  was  also  proved  in  §  262. 
The  proof  just  given  is  due  to  Bhaskara 
(born  1114  a.d.),  who  constructed  the 
"  Behold ! " 


16 


8P3 


16 


-T!\ 

^-=^      6  ^  \ 

\                1A           A  V- 

W        \6    iP_    M 

^   ^-^''" 

T" 

± 

figure 


simply 


AREAS   OF  POLYGONS. 


161 


319.  Theokem.  The  area  of  the  square  described  on 
the  hypotenuse  of  a  right  triangle  is  equal  to  the  sum  of 
the  areas  of  the  squares  described  on  the  other  two  sides. 


n 


Given  the  right  triangle  ABC,  and  the  squares  I,  II,  and  III  con- 
structed as  shown  in  the  figure. 

To  prove  that  D I  =  D II  4-  □  HI. 

Proof  :    Complete  the  rectangle  ABCE. 

Construct  A  GFH^  A  ACE. 

Draw  EH  and  produce  AE  to  meet  the  line  FH  at  K. 

Prove  that  (a)  AAEC^A  HEX. 

(5)  AEHG  is  a  parallelogram  whose  base  AE  and  alti- 
tude HK  are  each  equal  to  a  side  of  D  III. 

(^)  ECFH  a  parallelogram  whose  base  EC  and  altitude 
EK  are  each  equal  to  a  side  of  D II. 

But  these  two  parallelograms  together  are  equivalent 
ton  I  (Why?). 

The  student  should  give  the  proof  iu  detail,  making  an  outline  of 
the  steps  and  showing  how  each  step  is  needed  for  the  next.  For  ex- 
ample, why  is  it  necessary  to  prove  AHEK  ^  A  A  EC. 

Compare  the  various  proofs  of  this  theorem  that  are  given  in  this 
book.  The  method  of  counting  squares  shown  on  the  opposite  page 
is  applicable  to  all  cases  where  the  two  sides  are  commensurable. 


162  PIANE  GEOMETRY. 

Historical  Note.  The  theorem  of  §  319  is  one  of  the  most  impor- 
tant in  all  mathematics.  It  is  now  fairly  certain  that  the  general 
theorem  was  first  stated  and  proved  by  Pythagoras,  though  the  story 
that  he  sacrificed  100  oxen  to  the  gods  on  the  occasion  may  be  ques- 
tioned. Special  cases  of  the  theorem  were  known  to  the  Egyptians 
as  early  as  2000  B.C.,  e.g.  that  a  triangle  whose  sides  are  3,  4,  5  is 
right-angled. 

In  this  connection  the  Pythagoreans  also  discovered  the  irrational 
number,  that  is,  that  there  are  numbers  such  as  V2  which  cannot  be 
expressed  exactly  as  integers  or  as  ordinary  fractions. 

320.  EXERCISES. 

1.  The  bases  of  a  trapezoid  are  8  and  12  inches,  and  its  altitude  is 
8  inches.     Find  its  area. 

2.  The  bases  of  a  trapezoid  are  16  and  20  inches,  respectively,  and 
the  area  of  the  smaller  of  its  component  triangles  is  80  square  inches. 
Find  the  area  of  the  trapezoid.     See  the  figure  in  §  152. 

3.  State  in  words  the  geometric  theorem  indicated  in  each  of  the 
following,  and  draw  a  figure  to  illustrate  each  case : 

(a)  h(a +b)  =  ah  +  bh. 

(6)  (a  +  by  +(a-  by  =  2  (a2  +  b^). 
(c)  (a -{*  by  -  (a  -  by  :^  i  ab. 

4.  Show  that  the  rectangle  whose  base  is  a  +  J  and  whose  altitude 
h  a  —  b  has  the  same  perimeter  as  the  square  whose  side  is  a.  By 
means  of  Ex.  2,  §  318,  compare  their  areas. 

5.  If  two  triangles  have  the  same  base  but  their  vertices  are  on 
opposite  sides  of  it,  and  if  the  segment  joining  their  vertices  is  bi- 
sected by  the  common  base,  extended  if  necessary,  then  the  two 
triangles  are  equivalent. 

6.  State  and  prove  the  converse  of  the  theorem  in  the  preceding 
exercise. 

7.  The  bases  of  a  parallelogram  lie  on  the  parallel  lines  l^  and  I^ 
A  triangle  whose  base  is  equal  to  that  of  the  parallelogram  has  its 
vertex  in  l^  and  its  base  in  l^.     Compare  their  areas. 

8.  Prove  that  all  triangles  having  the  same  vertex  and  equal  bases 
lying  in  the  same  straight  line  are  equal  in  area. 


AREAS   OF  POLYGONS. 


163 


321.  Theorem.  The  areas  of  similar  triangles  are  in 
the  same  ratio  as  the  squares  of  any  two  corresponding 
sides,  or  as  the  squares  of  any  two  corresponding  altitudes. 


A  C   D  A'  C   D' 

Given  A  ABC  and  A'BC,  with  altitudes  BD  and  B'l>. 
To  prove  that 

area  A  ABC  AB^         BC?         CA?  B^ 


Proof: 
But 

Hence 


Give  the  proof  for  each  of  the  other  pairs  of  corre- 
sponding sides. 

322.  EXERCISE. 

Verify  the  preceding  theorem  by  counting  triangles  in  the  accom- 
panying figure. 


kyq^aa'b'c'     a'b'^     b'c'^     c'a'^ 

B'D'"" 

area,  A  ABC    _    ^(AC  x  BD)    _ 

AC          BD 

area  Aa'b'c'      ^a' c'  x  B^D') 

A'C'       B'D' 

AC         BD 
A'&       B'D' 

Why? 

are'dAABC^       A(/         BD^ 

'dTesi  aa'b'c'     a'c'^    b'd'^ 

Why? 

164 


PLANE  GEOMETRY, 


323.  Theorem.  The  areas  of  two  similar  polygons 
are  in  the  same  ratio  as  the  squares  of  any  two  corre- 
sponding sides  or  any  two  corresponding  diagonals, 

B' 


E  E' 

Outline  of  Proof:   Let  I,  II,  •••,  T,  11', 
areas  of  pairs  of  similar  triangles. 

r    ir    iir    iv 

I  +11  4- III  4- IV  ^  I 
V 


Show  (1) 
(2) 
(3) 


stand  for  the 
(§271.) 

^-  (By  §321.) 


r  +  ir  +  iir  +  iv' 

ABCDEF      _  I 


AB 


(By  §  268.) 
Bd^ 


a'b'c'd'e'f' 
Give  the  proof  in  detail. 


¥^'^ 


324. 


EXERCISES. 


1.  Verify  the  above  theorem  by  counting  the  number  of  equal 
hexagons  into  which  these 
two  similar  hexagons  are 
divided,  and  also  taking  the 
square  of  the  length  of  one 
side  in  each,  using  a  side  of  a 
small  hexagon  as  a  unit. 

2.  A  certain  triangular 
field  containing  2  acres  is  10 
rods  long  on  one  side.  Find 
the  area  of  a  similar  triangular  field  whose  corresponding  side  is  50  rods. 


AREAS   OF  POLYGONS.  165 

3.  The  areas  of  two  similar  triangular  flower  beds  are  24  square 
feet  and  36  square  feet  respectively.  If  a  side  of  one  bed  is  8  feet,  find 
the  corresponding  side  of  the  other. 

4.  If  similar  polygons  are  constructed  on  the  three  sides  of  a  right 
triangle,  show  that  the  one  described  on  the  hypotenuse  is  equivalent 
to  the  sum  of  the  other  two. 

Suggestions.  Let  a,  b,  c  represent  the  two  legs  and  hypotenuse  of 
the  triangle  and  A^B,C  the  areas  of  the  corresponding  polygons.  Then 
A      a2  B     &2 

—  =  — and  — =  — . 
C      c'^  C      c^ 

Hence  ^^^^  =  ^^^^  =  ^  =  1  ov  A -\- B  =  C.    Give  all  reasons  in 
C  c2         6-2 

full. 

5.  Find  a  line-segment  such  that  the  equilateral  triangle  described 
upon  it  has  four  times  the  area  of  the  equilateral  triangle  whose  side 
is  3  inches  long. 

6.  Show  that  the  square  on  the  altitude  of  an  equilateral  triangle 
is  three  fourths  the  square  on  a  side. 

7.  If  in  a  right  triangle  a  perpendicular  is  let  fall  from  the  vertex 
of  the  right  angle  to  the  hypotenuse,  show  that  the  areas  of  the  two 
triangles  thus  formed  are  in  Uie  same  ratio  as  the  adjacent  segments 
of  the  hypotenuse,  and  also  as  the  squares  of  the  adjacent  sides  of  the 
triangle. 

8.  Draw  a  line  from  a  vertex  of  a  triangle  to  a  point  in  the  oppo- 
site side  which  shall  divide  the  triangle  into  two  triangles  whose  ratio 
is  2:  5.     Also  2:1. 

9.  Divide  a  parallelogram  into  three  equivalent  parts  by  lines 
drawn  from  one  vertex.     Use  the  last  construction  in  Ex.  8. 

10.  The  sides  of  two  equilateral  triangles  are  8  and  6  respectively. 
Find  the  side  of  an  equilateral  triangle  whose  area  shall  be  equivalent 
to  their  sum.     Use  the  result  in  Ex.  4. 

11.  State  and  solve  a  problem  like  the  preceding  for  the  difference 
of  the  areas. 

12.  Corresponding  sides  of  two  similar  triangles  are  a  and  h. 
Find  the  side  of  a  third  triangle  similar  to  these  whose  area  is  equal 
to  the  sum  of  their  areas. 

13.  Likewise  for  any  two  similar  polygons. 


> 


166  '  PLANE  GEOMETRY. 

CONSTRUCTIONS. 

The  theorems  of  this  chapter  lead  to  numerous  construc- 
tions of  practical  importance. 

325.  Problem.  To  construct  a  square  equivalent  to 
the  sum  of  two  or  more  given  squares. 

Construction.  In  the  case  of  two  given  squares,  con- 
struct a  right  triangle  whose  legs  are  sides  of  the  two 
given  squared.     How  is  the  desired  square  obtained  ? 

In  the  case  of  three  given  squares,  use  the  square  result- 
ing from  the  first  construction  together  with  the  third 
square,  and  so  on. 

Give  the  construction  and  proof  in  full. 

326.  Problem.  To  construct  a  square  equivalent  to 
a  given  rectangle. 

Construction.  If  the  base  and  altitude  of  the  rectangle 
are  a  and  b  respectively,  we  seek  the  side  s  of  a  square 
such  that  8^=  ab ;  that  is,  we  s^ek  a  mean  proportional 
between  a  and  b.     See  Ex.  4,  §  295. 

Give  the  construction  and  proof  in  full  here. 

327.  Problem.  To  construct  a  square  equivalent  to  a 
given  triangle. 

Construction.  Show  how  to  modify  the  preceding  con- 
struction to  suit  this  case. 

328.  EXERCISES. 

Show  how  to  construct  each  of  the  following : 
•  1.    A  square  equivalent  to  the  difference  of  two  given  squares. 

2.  A  square  equivalent  to  the  sum  of  two  given  rectangles. 

3.  A  square  equivalent  to  the  sum  of  two  given  triangles. 

4.  A  square  equivalent  to  the  difference  of  two  given  rectangles, 
also  to  the  difference  of  two  given  triangles. 


AREAS   OF  POLYGONS. 


167 


329.   Problem.     To  construct  a  triangle  equivalent  to 
a  given  polygon,  c 

D 


Given  the  polygon  ABODE. 

To  construct  the  triangle  PCO  equivalent  to  ABODE. 

Construction.     Cut  off  A  ABC  by  the  segment  AC. 

Through  B  draw  BP  II  AC  to  meet  EA  extended  at  P. 

Draw  OP. 

In  a  similar  manner  draw  00. 

Then  PCO  is  the  required  triangle. 

Proof  :     PCDE  =  ABODE  since  A  APO  =  A  ABC  (Why  ?). 

Further,  PCDE  has  one  less  vertex  than  ABODE. 

But  A  PCO  =  PCDE  since  A  ECD  =  A  ECO  (Why?). 

Hence,  A  PCO  is  equivalent  to  the  given  polygon. 

330.    Problem.     To  construct  a  rectangle  on  a  given 
base  and  equivalent  to  a  given  parallelogram. 


b' 


Construction.  Let  b  and  h  be  the  base 
and  altitude  of  the  given  parallelogram,  b' 
the  base  of  the  required  rectangle,  and  x 
the  unknown  altitude. 

Then  we  are  to  determine  x  so  that 
b'x  =  hh,  that  is^  h'  :  h  =  h  :  x. 

Hence,  x  is  the  fourth  proportional  to  5',  b,  and  h. 
(See  §  291.)  Construct  this  fourth  proportional,  showing 
the  complete  solution. 

This  construction  is  attributed  to  Pythagoras.  It  represents  a 
much  higher  achievement  than  the . discovery  of  the  Pythagorean 
proposition  itself. 


168  PLANE  GEOMETRY. 

•% 
^  331.  EXERCISES. 

1.  Show  how  to  modify  the  last  construction  in  case  the  given 
figure  is  a  triangle.     Give  the  construction. 

2.  Construct  a  rectangle  on  a  given  base  equivalent  to  a  given 
irregular  quadrilateral. 

3.  Construct  a  rectangle  on  a  given  base  equivalent  to  an  irregu- 
lar hexagon. 

4.  On  a  side  of  a  regular  hexagon  as  a  base  construct  a  rectangle 
equivalent  to  the  hexagon. 

5.  Construct  a  parallelogram  on  a  given  base  equivalent  to  a  given 
triangle.     Is  there  more  than  one  solution? 

6.  Construct  a  square  whose  area  shall  be  three  times  that  of 
a  given  square;  five  times.     One  half  the  area;  one  fifth. 

7.  Construct  an  isosceles  triangle,  with  a  given  altitude  h,  equiva- 
lent to  a  given  triangle. 

8.  Draw  a  line  parallel  to  the  base  of  a  triangle  and  cutting  two 
of  its  sides.  How  will  the  resulting  triangle  and  trapezoid  compare 
in  area, 

(a)  If  each  of  the  two  sides  of  the  triangle  is  bisected? 
(h)  If  each  of  the  two   sides  of  the  triangle  is  three  times  the 
length  of  the  corresponding  side  of  the  trapezoid  ? 

9.  Construct  a  triangle  whose  base  and  altitude  are  equal  and 
whose  area  is  equal  to  that  of  a  given  triangle. 

10.  In  a  parallelogram  ABCD,  any  point  E  on  the  diagonal  BD  is 
joined  to  A  and  C.  Prove  that  A  BE  A  and  BEC  are  equivalent,  and 
also  that  A  DEA  and  DEC  are  equivalent. 

11.  The  sides  of  a  triangle  are  6,  8,  9.  A  line  parallel  to  the 
longest  side  divides  the  triangle  into  a  trapezoid  and  a  triangle  of 
equal  areas.     Find  the  ratio  in  which  the  line  divides  the  two  sides. 

12.  Draw  a  line  parallel  to  the  base  of  any  triangle,  and  cutting  two 
of  its  sides.  How  do  the  altitudes  of  the  resulting  triangle  and  trape- 
zoid compare, 

(a)  If  they  are  equal  in  area  ? 

(b)  If  the  area  of  the  triangle  is  three  times  that  of  the  trapezoid? 


AREAS   OF  POLYGONS.  169 

13.  Through  a  point  on  a  side  of  a  triangle  draw 
a  line  dividing  it  into  two  equivalent  parts. 

Solution.  Let  P  be  the  given  point.  Draw  the 
median  BD.  Draw  BE  ||  PB  and  draw  PE.  This  is 
the  required  line.     Prove. 

Suggestion.  Notice  that  ADPE  =  ADPB  by 
Ex.  4,  §  313,  and  hence  that  AEOD  =  ABOP. 

14.  Through  a  given  point  on  a  triangle  draw  a  line  which  divides 
it  into  two  figures  whose  areas  are  in  the  ratio  i. 

15.  Inscribe  a  circle  in  a  triangle,  touching  its 
sides  in  the  points  D,  E,  F.  With  the  vertices 
as  centers,  construct  circles  passing  through  these 
points  in  pairs.  Show  that  each  of  these  latter 
circles  is  tangent  to  the  other  two. 

SUMMARY  OP  CHAPTER  IV. 

1.  State  what  is  meant  by   the  area  of  a  rectangle.     Give  the 
formula. 

2.  Give  formulas  for  areas  of  parallelograms  and  triangles. 

3.  How  is  the  formula  for  the  area  of  a  trapezoid  obtained? 

4.  What  theorems  of  this  chapter  can  be  stated  algebraically,  as 
(a  +  by  =  a^  +  2ab-\-b^ 

5.  State  the  theorem  on  the  ratio  of  the  areas  of  two  similar  tri- 
angles; two  similar  polygons.     Give  examples. 

6.  Tabulate  the  problems  of  construction  given  in  this  chapter. 

7.  If  two  rectangles  have  the  same  base,  how  does  the  ratio  of 
their  areas  compare  with  the  ratio  of  their  altitudes? 

8.  If  two  triangles  have  equal  altitudes,  how  does  the  ratio  of  their 
areas  compare  with  the  ratio  of  their  bases? 

9.  State   all  theorems  of  this  chapter  proved  *by  means  of  the 
Pythagorean  proposition. 

10.  State  some  of  the  more  important  applications  of  the  theorems 
in  this  chapter.  Return  to  this  question  after  studying  the  succeeding 
list  of  problems. 


170  PLANE  GEOMETRY. 

PROBLEMS  AND  APPLICATIONS. 

1.  Find  the  area  of  a  square  whose  diagonal  is  6  inches. 

2.  Find  the  area  of  a  square  whose  diagonal  is  d  inches. 

3.  ABCD  is  a  square  placed  at  the  crossing  of  two  strips  of  equal 
width,  as  shown  in   the   figure.     The   small  black 
square  has  two  vertices  on  the  sides  of  the  horizontal 
strips  and  two  on  the  sides  of  the  vertical  strip. 

(a)    Find  the  area  of  each  square  when  the  width 
of  the  strips  is  4  inches. 

(6)    Compare  the  area  of  the  black  square  and 
the  white  border  surrounding  it. 

(c)  Can  squares  be  placed  as  in  the  figure  in  case  the  strips  are  of 
unequal  width?  In  the  two  following  questions  let  the  small  square^ 
be  drawn  with  two  vertices  on  the  sides  of  the  horizontal  strip  and 
one  diagonal  parallel  to  these  sides. 

(jl)  If  the  horizontal  strip  is  4  inches  wide,  what  must  be  the 
width  of  the  vertical  strip  in  order  that  the  large  square  may  have 
twice  the  area  of  the  small  one? 

Hint.     The  diagonal  of  the  small  square  is  4  inches. 

(e)  If  the  horizontal  strip  is  a  inches  wide,  what  must  be  the 
width  of  the  vertical  strip  in  order  that  the  area  of  the  black  square 

shall  be  -  the  area  of  the  larger  square  ? 
n 

4.  Prove  that  the  area  of  a  rhombus  is  one  half  the  product  of  its 
diagonals. 

5.  Prove  that  the  area  of  an  isosceles  right  triangle  is  equal  to 
the  square  on  the  altitude  let  fall  upon  the  hypotenuse. 

6.  If  the  diagonals  of  a  quadrilateral  intersect  at  right  angles, 
prove  that  the  sum  of  the  squares  on  one  pair  of  opposite  sides  is 
equal  to  the  sum  of  the  squares  on  the  other  two  sides. 

7.  Inscribe  a  square  in  a  semicircle  and  in  a  quadrant  of  the 
same  circle.     Compare  their  areas.     See  Ex.  8,  page  147. 

8.  In  the  triangle  ABC,  CD  is  an  altitude.  E  is 
any  point  on  CD.  If  DE  is  one  half  CD,  compare  the 
area  of  the  triangle  AEB  and  the  sum  of  the  areas  of 
the  triangles  AEC  and  BEC.  Also  compare  these  areas 
if  DE  is  one  nth  oi  DC. 


AREAS   OF  POLYGONS, 


171 


9.    A  BCD  is  a  square,  and  E,  F,  G,  H  are  the  middle  points  of  its 
sides.     On  EF,  points  N  and  Q  are  taken  so  that 
PN  =  PQ.     Similarly  KL  =  KM,  also  KL  =  NP, 
and  so  on. 

(a)    Prove  that  .O/OL  is  a  rhombus. 

(h)  If  AB=Q  inches  and  if  KM  =  ME,  find 
the  sum  of  the  areas  of  the  four  rhombuses  AMOL, 
BQON,  etc. 

(c)  If  HE  =  8  inches,  what  is  the  area  of  the 
whole  square  ? 

{(l)  What  part  of  KE  must  KM  be  in  order  that 
the  sum  of  the  rhombuses  shall  be  \  the  area  of  the 
square? 

(g)    If  AB 


a,  find  KM  so  that  the  sum  of  the 


rhombuses  shall  be  -  the  area  of  the  square. 


Parquet  Flooring. 


(/)    Prove  that  L,  0,  R  lie  in  the  same  straight  line. 

10.  ABCD  is  a  square  and  E,  F,  G,  Hare  the  middle  points  of  its 
sides.     SN  =  PK=QL  =  RM. 

(a)  li  AB  =  Q  inches  and  ii  PK  =  I  inch,  find 
the  sum  of  the  areas  of  the  triangles  EHN,  EFK, 
FGL,  and  GHM. 

*  (&)  li  AB  =  6,  find  PK  so  that  the  sum  of  the 
four  triangles  EFK,  etc.,  shall  be  one  fifth  of  the 
whole  square. 

,  {c)  li  AB  =  Q  and  if  the  points  E,  K,  Q  lie  in 
(a  straight  line,  find  the  sum  of  the  areas  of  these 
triangles. 

(d)  If  AB  =  a  and  if  PK  =  one  nth  of  PO, 
/^  find  the  sum  of  the  areas  of  the  triangles. 

(g)  If  AB  =  a,  find  PK  so  that  the  sum  of  the 
triangles  shall  be  one  mih  of  the  whole   square. 
What  will  be  the  length  of  PK  in  case  the  tri-      Parquet  Flooring, 
angles  occupy  one  half  of  the  whole  square? 

11.  The  sides  of  two  equilateral  triangles  are  a  and  h  respectively. 
Find  the  side  of  an  equilateral  triangle  whose  area  is  equal  to  the  sum 
of  their  areas. 


PLANE  GEOMETRY, 


lip, 

/ 

12.  Construct  a  triangle  similar  to  a  given  triangle  and  having  16 
times  the  area. 

13.  The  middle  points  of  the  sides  of  a  quadrilateral  are  con- 
nected. Show  that  the  area  of  the  parallelogram  so  formed  is  half  the 
area  of  the  quadrilateral. 

14.  ABCD  is  a  square.  Each  side  is  divided 
into  four  equal  parts  and  the  construction  com- 
pleted as  shown  in  the  figure. 

(a)  Prove  that  QNOH,  KLOE,  etc.,  are  par- 
allelograms. 

(b)  What  part  of  the  area  of  the  square  is  oc- 
cupied by  these  four  parallelograms? 

(c)  What  part  of  the  area  of  the  square  is  oc- 
cupied by  the  four  triangles  NEC,  LGO,  PFO, 
andilfi^O? 

{d)  If  AB  =  6,  find  the  lengths  of  KO  and 
QP. 

(e)  Find  the  ratio  of  the  segments  KO  and 
QP.  Does  this  ratio  depend  upon  the  length  of 
ABl 

{/)  In  the  parquet  floor  design  what  fraction  is  made  of  the  dark 
wood?     Does  this  depend  upon  the  size  of  the  original  square? 

15.  On  a  given  line-segment  AB  as  a  hypotenuse  construct  a  right 
triangle  such  that  the  altitude  upon  the  hypotenuse  shall  meet  it  at  a 
given  point  D. 

16.  If  ABC  is  a  right  triangle  and  CD±AB, 


Parquet  Flooring. 


prove  that 


AC" 
CB' 


AD 
DB 


17.  By  means  of  Exs.  15  and  16  construct  two  segments  HK  and 
LM  such  that  the  ratio  of  the  squares  on  these  segments  shall  equal  a 
given  ratio. 

18.  Divide  a  given  segment  into  two  segments  such  that  the  areas 
of  the  squares  constructed  upon  them  shall  be  in  a  given  ratio. 

ATB 
f    19.   On  a  given  segment  A  B  find  a  point  D  such  that   ■  =  2. 

Ajf 

20.    Construct  a  line  parallel  to  the  base  of  a  triangle  such  that  the 

resulting  triangle  and  trapezoid  shall  be  equivalent. 


AREAS  OF  POLYGONS. 


173 


21.    Construct  two  lines  parallel  to  the  base  of  a  triangle  so  that 
the  resulting  two  trapezoids  and  the  triangle  shall  have  equal  areas. 


^^r  ^ 


D 


^  22.  In  each  of  the  accompanying  de- 
signs for  tile  flooring  find  what  fraction 
of  the  space  between  the  lines  AB  and 
CD  is  occupied  by  tiles  of  eacji  color. 

Study  each  design  with  care  to 
see  that  the  character  of  the  figure 
determines  the  relative  sizes  of  the 
various  pieces  of  tile. 

I        23.   ABCD  is  a  square.     Points  E,  F,  G,  •••  are  so   taken  that 
AE  =  AF=  GB  =  BH=  -•■  . 

(a)  If  AB  =  6  and  .4F  =  1,  find  the  sum  of  the      D_M_ 
areas    of   the    four   triangles   EFO,    GHO,   KLO,     j^^ 
MNO. 

(b)  Find  the  sum  of  these  areas  if  AB  =  a  and 
AF=h. 

(c)  If  AB  =  Q  and  if  the  sum  of  the  areas  of      ^    ^ 
the  triangles  is  9  square  inches,  find  A  F. 

(d)  li  AB  =  a  and  if  the  sum  of  the  areas  of  the  triangles  is  — ,  find 

n 

\y  A  F.     Interpret  the  two  results. 

_     24.    Show  how  to  construct  a  square  whose  area  is  n  times  the  area 
of  a  given  square. 

25.   Construct  a  triangle  similar  to  a  given  triangle  and  equivalent 
to  n  times  its  area. 

^     26.   Construct  a  hexagon  similar  to  a  given  hexagon  and  equivalent 
''    to  n  times  its  area. 

27.    Show  how  to  construct  a  polygon  similar  to  a  given  polygon 
and  equivalent  to  n  times  its  area. 


174 


PLANE  GEOMETRY. 


28.    The  alternate  middle  points  of  the  sides  of  a  regular  hexagon 
are  joined  as  shown  in  the  figure. 

(a)  Are  the  triangles  thus  formed  equilateral  ?     Prove. 

(b)  Is  the  star  regular  (i.e.   are  its  six  acute  angles  equal  and  its 
sides  equal)  ? 

(c)  Compare  the  three  segments  into  which  each  triangle  divides 
the  sides  of  the  other. 

(d)  Is  the  inner  hexagon  regular?    Prove.     See  Ex.  1,  p.  76. 


^^H 

I: 

III 

MH  ~ 

Tile  Pattern. 

\  (e)  If  AB  =  6  in.,  find  the  area  of  the  large  hexagon,  the  star,  and 
the  small  hexagon. 
V,     29.   A  border  is  to  be  constructed  about 
a  given  square  with  an  area  equal  to  one 
half  that  of  the  square. 

(a)  By  geometrical  construction  find  the 
outer  side  of  the  border  if  the  side  of  the 
square  is  given. 

(b)  If  an  outer  side  of  the  border  is  24 
in.,  find  a  side  of  the  square,  its  area  being 
two  thirds  that  of  the  border. 

\ .  30.  A  border  is  constructed  about  a 
given  regular  octagon,  such  that  its  area  is 
equal  to  that  of  the  octagon. 

(a)  If  a  side  of  the  given  octagon  is  a  , 
given   segment   AB,   find  by  geometrical 
construction  a  segment  equal  to  an  outer 
side  of  the  border. 

(b)  If  a  side  of  the  given  octagon 
is  16  in.,  find  an  outer  side  of  the 
border.  Ceiling  Pattern. 


Ceiling  Pattern. 


AREAS   OF  POLYGONS. 


175 


J 


n  /\c  A     / 

r  IX  X 

I'x; 

() 

^' \m ■< 

'  t 

31.  The  accompanying  design  is  based  on 
a  set  of  squares  such  as  ABCD.  The  small 
triangles  are  equal  isosceles  right  triangles 
constructed  as  shown. 

(a)  Are  the  vertical   and  the  horizontal 
sides  of  the  octagons  equal  ? 
(6)  Are  the  octagons  regular  ? 

(c)  If  a  side  of  one  of  the  squares  is  6,  find 
the  area  of  one  of  the  octagons. 

(d)  What  fraction  of  the  whole  tile  design 
is  occupied  by  the  light-colored  tiles?  Does 
this  depend  upon  the  size  of  the  original 
squares  ? 

32.  Given  two  lines  at  right  angles  to  each 
other.  Find  the  locus  of  all  points  such  that 
the  sum  of  the  squares  of  the  distances  from  the  lines  is  25. 

33.  Given  two  concentric  circles  whose  radii  are  r  and  r'.  Find 
the  length  of  a  chord  of  the  greater  which  is  tangent  to  the 
smaller. 

34.  If  two  equal  circles  of  radius  r  intersect  so  that  each  passes 
through  the  center  of  the  other,  find  the  length  of  the  common 
chord. 

35.  The  square  on  the  hypotenuse  of  a  right  triangle  is  four  times 
the  square  on  the  altitude  upon  the  hypotenuse.     Prove  it  isosceles. 

36.  In  a  right  triangle  the  hypotenuse  is  10  feet  and  the  difference 
between  the  other  sides  is  2  feet.     Find  the  sides. 

37.  Two  equal  circles  are  tangent  to  each  other  and  each  circle  is 
tangent  to  one  of  two  lines  perpendicular  to  each  other.  Find  the 
locus  of  the  points  of  tangency  of  the  two  circles. 

Suggestion.  Note  that  the  point  of  tangency  bisects  their  line  of 
centers  and  that  the  centers  move  along  lines  at  right  angles  to  each 
other. 

38.  The  square  on  a  diagonal  of  a  rectangle  is  equal  to  half  the 
sum  of  the  squares  on  the  diagonals  of  the  squares  constructed  on 
two  adjacent  sides  of  the  rectangle. 

39.  Show  that  the  diagonals  of  a  trapezoid  form  with'  the  non- 
parallel  sides  two  triangles  having  equal  areas. 


CHAPTER   V. 

REGULAR  POLYGONS  AND  CIRCLES. 
REGULAR  POLYGONS. 

332.  A  regular  polygon  is  one  which  is  both  equilateral 
and  equiangular. 

According  to  this  definition,  determine  whether  each  of 
the  following  polygons  is  regular  or  not  and  state  why  : 

An  equilateral  triangle,  an  equiangular  triangle,  a  rectangle,  a 
square,  a  rhombus.  Draw  a  figure  to  illustrate  each.  Make  a  triangle 
which  fulfills  neither  condition  of  the  definition,  also  a  quadrilateral. 

333.  The  general  problem  of  constructing  a  regular 
polygon  depends  upon  the  division  of  a  circle  into  as  many 
equal  parts  as  the  polygon  has  sides. 

The  problem  of  dividing  the  circle  into  equal  parts  can 
be  solved  in  some  cases  by  the  methods  of  elementary 
geometry,  and  some  of  these  methods  will  be  considered 
in  this  chapter.  In  most  cases  this  problem  cannot  be 
solved  by  elementary  methods. 

E.g.  the  circle  may  be  divided  into  2,  3,  4,  5,  6,  8,  10,  12,  15,  equal 
parts,  but  not  into  7,  9,  11,  13,  14,  equal  parts. 

If  a  circle  has  already  been  divided  into  a  certain  num- 
ber of  equal  parts,  it  may  then  be  divided  into  twice,  four 
times,  eight  times,  etc.,  that  number  of  parts  by  repeated 
bisection  of  the  arcs.     (See  §  204,  Ex.  1.) 

The  division  of  the  circle  into  equal  parts  depends  upon 
the  theorem  (§199)  that  equal  central  angles    intercept 

176 


REGULAR  POLYGONS  AND   CIRCLES,  177 

equal  arcs  on  the  circle,  and  hence  it  involves  the  subdi- 
vision of  angles  into  equal  parts. 

The  following  exercises  depend  upon  principles  already 
familiar. 

With  each  solution  give  the  reasons  in  full  for  each  step. 

All  constructions  are  to  be  made  with  ruler  and  com- 
passes only. 

334.  EXERCISES. 

1.  Divide  a  given  angle  or  arc  into  four  equal  parts. 

2.  Divide  a  given  angle  or  arc  into  eight  equal  parts. 

3.  If  an  angle  or  arc  is  already  divided  into  a  certain  number  of 
equal  parts,  show  how  to  divide  it  into  twice  that  number  of  equal 
parts. 

4.  Divide  a  circle  into  four  equal  parts. 

5.  Divide  a  circle  into  eight  equal  parts. 

6.  If  a  circle  is  already  divided  into  a  certain  number  of  equal 
parts,  show  how  to  divide  it  into  twice  that  number  of  equal  parts. 

7.  Divide  a  circle  into  six  equal  parts. 
Suggestion.     Construct  at  the  center  an  angle  of  60°. 

8.  Draw  the  chords  connecting  in  order  the  four  division  points 
in  Ex.  4,  and  show  that  the  figure  is  a  regular  quadrilateral. 

9.  Draw  the  tangents  at  the  four  division  points  in  Ex.  4,  and 
show  that  a  regular  quadrilateral  is  formed. 

10.  Draw  the  chords  connecting  in  order  the  division  points  in 
Ex.  7,  and  show  that  a  regular  hexagon  is  formed.  Prove  that  the 
side  of  the  hexagon  is  equal  to  the  radius  of  the  circle. 

11.  Draw  chords  connecting  alternate  division  points  in  Ex.  7,  and 
show  that  a  regular  triangle  is  formed. 

12.  Construct  tangents  to  the  circle  at  alternate  division  points  in 
Ex.  7,  and  show  that  a  regular  triangle  is  formed. 

13.  Draw  tangents  to  the  circle  at  the  division  points  in  Ex.  7,  and 
show  that  a  regular  Jiexagon  is  formed. 

Note.  See  also  the  construction  of  regular  polygons  of  3,  4,  and 
6  sides,  §  163,  Exs.  3,  4,  5. 


178  PLANE  GEOMETRY. 

335.  Theorem.  If  a  circle  is  divided  into  any  num- 
her  of  equal  arcs,  the  chords  joini7ig  the  division  points, 
taken  in  order,  form  a  regidar  polygon. 

Proof  :  Show  (1)  that  the  chords  are  equal ;  (2)  that 
the  angles  are  inscribed  in  equal  arcs  and  hence  are  equal. 

336.  Theorem.  If  a  circle  is  divided  into  any  num- 
ber of  equal  parts,  the  tangents  at  the  points  of  divi- 
sion, taken  in  order,  form  a  regular  polygon. 


Given  the  G  OA  divided  into  equal  arcs  by  the  points  M,  N,  P,  etc., 
with  tangents  drawn  at  these  points  forming  the  polygon  ABCDEF. 

To  prove  that  ABCDEF  is  a  regular  polygon. 

Analysis  of  Proof :    (1)  To  prove  that  AB  =  BC=^  CD,  etc. 

We  know  that  BP  =  BN  (Why  ?). 

Hence,  if  we  can  show  that  AN  =  NB  and  BP  =  PC,  then 
it  will  follow  that  AB  —  BC. 

To  prove  AN  =  NB,  it  must  be  shown  that  Z 1  =  Z  2, 
and  this  is  done  by  showing  that  Z 1  and  Z  2  are  halves  of 
the  equal  angles  NOM  and  NOP. 

This  necessitates  proving  A  ^OiV  ^  A  AOM. 

Now  state  the  proof  in  full. 

(2)  To  prove  ZABC  =  Zbcd  =  Z  CDE,  etc. 

From  the  triangles  proved  congruent  under  (1)  show 


REGULAR   POLYGONS  AND   CIRCLES. 


179 


that  Z MAO  =  Z  OAN  =  Z  NHO  =  ZoBP;  and  hence, 

Z  MAO-\-  Z  OAN  =  Z  NBO  +  Z  07^7^  or  Z  i^Jl?  =  Z  ABC. 

In  like  manner  it  is  proved  that  Z^7:?C=  Z  BCD^  etc. 

Hence  the  polygon  is  equilateral  and  also  equiangular, 
and  hence  regular. 

337.  Theorem.     If  a  polygon  is  regular,  a  cwcle  may 
he  circumscribed  about  it. 

D 


A  B 

Given  a  regular  polygon  ABODE. 
To  prove  that  a  point  o  can  be  found  such  that 

OA  =  OB  —  OC  —  0B=  OE. 
Outline  of  Proof :    Bisect  A  A  and  B  and  let  the  bisectors 
meet  in  some  point  O.     Then  o  is  the  center  sought. 
For  we  have     (1)         OA  =  OB,  Why  ? 

(2)  A  AOB  ^  BOC.      .',  0A=  OC. 
(S)  A BOC  ^  A  COD.      .'.OB=OD. 
(4)  Now  prove  OC  =  OE. 
.  •.  OA  =  0B=  OC  =  OD  =  OE. 

Prove  each  step,  showing  how  it  depends  upon  the  pre-' 
ceding. 

338.  EXERCISES. 

1.  Find  the  locus  of  the  vertices  of  all  regular  polygons  of  the 
same  number  of  sides  which  can  be  circumscribed  about  the  same 
circle. 


180  PLANE  GEOMETRY. 

2.  Find  the  locus  of  the  middle  points  of  the  sides  of  all  regular  poly- 
gons of  the  same  number  of  sides  which  are  inscribed  iu  the  same  circle. 

3.  Is  an  equilateral  circumscribed  polygon  regular?     Prove. 

4.  Is  an  equiangular  circumscribed  polygon  regular?     Prove. 

339.  Theorem.  //  a  ^polygon  is  regular,  a  circle  may 
he  inscribed  in  it. 

Given  the  regular  polygon  ABCDE. 

To  prove  that  a  point  O  may  be  found  such  that  the  per- 
pendiculars   OM,   ON,  OP,  OQ,  OR, 
are  equal.  /x 

Outline  of  Proof  :    Determine  a  y<  >\^ 

point  O  as  in  the  proof  of  tlie  pre-    j^^^      \      /      ^^^c 
ceding    theorem.       Then    A  AOB,      \  '§--''''    / 

HOC,  etc.,  are  equal  isosceles  trian-       /A''' '/  ''\~    ~^~/v 
gles,  and  their  altitudes  are  equal.  \      /''    j      \      / 

Hence  O  is  the  required  point.  \/^        |        '\] 

Give  proof  in  full.  a        M         B 

340.  Theorem.  An  equilateral  polygon  inscribed  in 
a  circle  is  regular. 

Suggestion  for  Proof  :  Use  the  fact  that  equal  chords 
subtend  equal  arcs,  and  apply  §  335. 

341.  EXERCISE. 

Show  that  the  inscribed  and  circumscribed  circles  of  a  regular 
polygon  have  the  same  center. 

342»  Definitions.  The  center  of  a  regular  polygon  is  the 
common  center  of  its  inscribed  and  circumscribed  circles. 

The  radius  of  a  regular  polygon  is  the  distance  from  the 
center  to  one  of  its  vertices. 

The  apothem  of  a  regular  polygon  is  the  perpendicular 
distance  from  its  center  to  a  side. 


REGULAR  POLYGONS  ANU   CIRCLES.  181 

343.  Theorem.  The  area  of  a  regular  polygon  is 
equal  to  half  the  product  of  its  apothem  and  perimeter. 

Suggestion  for  Proof :  A  regular  polygon  is  divided 
by  its  radii  into  a  series  of  congruent  triangles,  the  area 
of  each  of  which  is  one  half  the  product  of  the  apothem 
and  a  side  of  the  polygon. 

Complete  the  proof. 

344.  Theorem.  The  area  of  any  polygon  circum- 
scribed about  a  circle  is  equal  to  one  half  the  product  of 
the  perimeter  of  the  polygon  and  the  radius  of  the  circle. 

The  proof  is  left  to  the  student. 

345.  EXERCISES. 

1.  Is  every  equiangular  polygon  inscribed  in  a  circle  regular? 
Prove. 

2.  Show  that  the  radius  of  a  regular  polygon  bisects  the  angle  at 
the  vertex  to  which  it  is  drawn. 

3.  Show  that  the  perimeter  of  a  regular  polygon  of  a  given  num- 
ber of  sides  is  less  than  that  of  one  having  twice  the  number  of  sides, 
both  being  inscribed  in  the  same  circle. 

4.  Show  that  the  perimeter  of  a  regular  polygon  is  greater  than 
that  of  one  having  twice  the  number  of  sides,  both  being  circum- 
scribed about  the  same  circle. 

5.  Compare  the  areas  of  the  two  polygons  in  p]x.  3. 

6.  Compare  the  areas  of  the  two  polygons  in  Ex.  4. 

$  7.  Show  that  the  area  of  a  square  inscribed  in  a  circle  of  radius 
R  is  2  R"^.  How  does  this  compare  with  the  area  of  the  circum- 
scribed square. 

\  8.  Compute  the  apothem  and  area  of  a  regular  inscribed  hexagon 
if  the  radius  of  the  circle  is  R ;  also  of  the  regular  circumscribed 
hexagon. 

9.  A  regular  triangle  is  inscribed  in  a  circle  of  radius  10.  Find 
the  apothem  and  a  side  of  the  triangle. 


182  PLANE  GEOMETRY, 

346.  Theorem.  Two  regular  polygons  of  the  same 
number  of  sides  are  similar. 

Outline  of  Proof:  (1)  Show  that  all  pairs  of  corre- 
sponding angles  are  equal. 

(2)  Show  that  the  ratios  of  pairs  of  corresponding  sides 
are  equal.     Hence  the  polygons  are  similar  (Why?). 

347.  Theorem.  The  perimeters  of  two  regular  poly- 
gons having  the  same  number  of  sides  are  in  the  same 
ratio  as  their  radii  or  their  apothems. 

Outline  of  Proof:  Show  (1)  that  each  triangle  formed 
by  a  side  and  two  radii  in  one  polygon  is  similar  to  the 
corresponding  triangle  in  the  other  polygon. 

A  Ti  f  d 

(T)  That  — --  =  —  =  — ,   where  AB  and  A'B'  are  the  two 
^  A^B^      r'      a' 

sides,  r  and  r'  the  corresponding  radii  and  a  and  a'  the 

corresponding  apothems.     And  so  for  the  remaining  pairs 

of  triangles. 

,gN    rp,     ,         AB  -{-  BC  -\-  CD  ^  •  •  •      _  AB   _r__o^ 

Draw  the  figure  and  give  the  proof  in  full. 

348.  Theorem.  The  areas  of  two  regular  polygons 
of  the  same  number  of  sides  are  in  the  same  ratio  as  the 
squares  of  the  corresponding  radii  or  apothems. 

Outline  of  Proof  :  Divide  the  polygons  into  pairs  of 
corresponding  triangles  as  in  the  preceding  proof. 

AT         r^         n^ 

(1)  Show  that  ,  =—,  =  -—,  and  so  for  each  pair  of 

A  r      r'^     a'^ 

AT  +  AlI-hATTI+  ...        r2 


triangles.  ^  ^ 


Draw  figure  and  give  the  proof  in  full. 


REGULAR   POLYGONS  AND   CIRCLES. 


PROBLEMS   AND   APPLICATIONS. 

1.  Find  the  ratio  of  the  perimeters  of  squares  inscribed  in  and 
circumscribed  about  the  same  circle. 

2.  Find  the  ratio  of  the  perimeters  of  regular  hexagons  inscribed 
in  and  circumscribed  about  the  same  circle. 

3.  Find  the  ratio  between  the  perimeters  of  regular  triangles 
inscribed  in  and  circumscribed  about  the  same  circle. 

4.  Find  the  ratio  of  the  areas  of  regular  triangles  inscribed  in  and 
circumscribed  about  the  same  circle.  Also  find  the  ratio  of  the  areas 
of  such  squares  and  of  such  hexagons. 

5.  The  perimeter  of  a  regular  hexagon  inscribed  in  a  circle  is 
24  in6hes.  Find  the  perimeter  of  a  regular  hexagon  circumscribed 
about  a  circle  of  twice  the  diameter. 

6.  The  area  of  a  regular  triangle  circumscribed  about  a  circle  is 
64  square  inches.  What  is  the  area  of  a  regular  triangle  inscribed  in 
a  circle  of  one  third  the  radius? 

7.  The  area  of  a  regular  hexagon  inscribed  in  a  circle  is  48  square 
inches.  What  is  the  area  of  a  regular  hexagon  circumscribed  about 
a  circle  whose  diameter  is  If  times  that  of  the  first? 

8.  A  chord  AB  bisects  the  radius  perpendicular  to  it.  Find  the 
central  angle  subtended  by  the  chord.     State  the  result  as  a  theorem. 

9.  State  and  prove  the  converse  of  the  theorem  in  Ex.  8. 

10.  Find  the  area  of  a  regular  triangle  inscribed  in  a  circle  whose 
radius  is  6  inches. 

mi  11.   Find  the  area  of  a  regular  triangle  inscribed  in  a  circle  whose 
radius  is  r  inches. 

12.   One  of  the  acute  angles  of  a  right  triangle  is  60°  and  the  side 
adjacent  to  this  angle  is  r  inches  long.     Find  the  remaining  sides  of 
the  triangle. 

,      13.    A  regular  triangle  is  circumscribed  about  a 
circle  of  radius  r.     Find  its  area. 

Suggestion.     First  find  DC  in  the  figure. 

4     14.   A  regular  triangle  of  area  36  square  inches  is 
inscribed  in  a  circle.     Find  the  radius  of  the  circle. 


184 


PLANE  GEOMETRY. 


/ 


15.  Find  the  radius  of  a  circle  if  the  area  of  its  regular  inscribed 
triangle  is  a. 

16.  Find  the  radius  of  the  circle  if  the  area  of  the  regular  circum- 
scribed triangle  is  a. 

17.  Find  the  radius  of  a  circle  if  the  difference  between  the 
perimeters  of  the  tegular  inscribed  and  circumscribed  triangles  is 
12  inches. 

18.  Find  the  radius  of  a  circle  if  the  difference  of,  the  perimeters 
of  the  regular  inscribed  and  circumscribed  hexagons  is  10  inches. 

19.  If  the  area  of  a  circumscribed  square  is  25  square  inches  greater 
than  that  of  an  inscribed  square,  what  is  the  diameter  of  the  circle  ? 

20.  Find  the  radius  of  a  circle  if  the  difference  between  the  areas 
of  the  inscribed  and  circumscribed  regular  triangles  is  25  square 
inches. 

21.  Find  the  radius  of  a  circle  if  the  difference  between  the  areas  of 
the  regular  inscribed  and  circumscribed  hexagons  is  25  square  inches. 

22.  The  difference  betwieen  the  areas  of  the  squares  circumscribed 
about  two  circles  is  50  square  inches  and  the  differ- 
ence  of  their  diameters  is  4   inches.     Find  each 
diameter. 

23.  If  the  inscribed  and  escribed  circles  0  and 
0'  of  an  equilateral  triangle  are  constructed  as 
shown  in  the  figure,  find  the  ratio  of  their  radii. 
Does  this  ratio  depend  upon  the  size  of  the  triangle  ? 

24.  Given  a  triangle  ABC  and  a  segment  a, 
show  how  to  construct  a  segment  DE  II  AB  and 
equal  to  the  segment  «,  such  that  the  points  D  and 
E  shall  lie  on  the  sides  CA  and  CB  respectively  or 
on  these  sides  extended. 

25.  A  triangular  plot  of  ground  ABC  is  to  be 
laid  out  as  a  triangular  flower  bed  with  a  walk  of 
uniform  width  extending  around  it. 

(a)  Prove  that  the  flower  bed  is  similar  to  the 
original  triangle. 

(ft)  Show  that  the  corners  of  the  flower  bed  lie  on 
the  bisectors  of  the  angles  of  the  original  triangle. 


REGULAR  POLYGONS  AND   CIRCLES.  185 

(c)  Find  a  segment  S  so  that  ''—;;-  =  2  and  construct  A'B'   equal 

to  S  and  parallel  to  AB,  the  points  A'  and  B'  lying  on  the  bisectors 
of  the  angles  A  and  B  respectively.     See  Exs.  16-19,  page  172. 

(rf)  Draw  A'C  II  AC  and  B'C  II  5C.  ' 

Prove  that  the  area  of  the  flower  bed  A' B'C,  as  thus  constructed, 
is  equal  to  the  area  of  the  walk. 

(e)  Construct  the  figure  for  the  flower  bed  so  that  its  area  is  five 
times  that  of  the  walk. 

26.  Given  a  rectangular  plot  of  ground.  Is  it  always  possible  to 
lay  off  on  it  a  walk  of  uniform  width  running  around  it  so  that  the 
plot  inside  the  walk  shall  be  similar  to  the  original  figure?     Prove.    . 

27.  Is  the  construction  proposed  in  Ex.  26  always  possible  in  the 
case  of  a  square  ?  of  a  rhombus  ?     Prove. 

28.  If  a  side  of  a  regular  hexagon  circumscribed  about  a  circle  is 
a,  find  tlie  radius  of  the  circle. 

29.  On  a  regular  hexagonal  plot  of  ground  whose  side  is  12  feet  a 
walk  of  uniform  width  is  to  be  laid  oft'  around  it.  Find  by  algebraic 
computation  the  width  of  the  walk  if  it  is  to  occupy  one  half  the 
whole  plot. 

30.  Find  by  geometric  construction  the  width  of  the  walk  in 
Ex.  29. 

31.  Show  that  the  figure  inside  the  walk  in  Ex.  29  is  a  regular 
hexagon. 

32.  Given  a  segment  AB,  find  three  points  C,  D,  E  on  it  so  that, 

AC^^l     A^^l    ^^^  AE'  ^  3 
AB^     4'    ab'     2'  ab'     4* 

33.  A  regular  hexagon  ABCDEF  is  to  be  divided  into  four  pieces 
of  equal  area  by  segments  drawn  parallel  to  its 

sides  forming  hexagons  as  shown  in  the  figure.  If 
AB  =  2^  feet,  find  a  side  of  each  of  the  other  hexa- 
gons and  also  the  apothem  of  each. 

34.  Without  computing  algebraically  the  apo- 
them or  sides  of  the  inner  hexagons  in  Ex.  33, 
show  how  to  construct  the  figure  geometrically. 
See  Ex.  6,  §  331.     Also  Ex.  18,  page  172.     Use  §348. 


C 


186 


PLANE  GEOMETRY, 


/ 


35.  In  a  given  hexagonal  polygon  whose  side  is  a  find  in  terms  of 
a.  and  n  the  width  of  a  walk  around  it  which  will  occupy  one  nth  of 
the  area  of  the  whole  polygon. 

36.  By  means  of  hexagons  similar  to  those  in  Ex.  33,  divide  a 
given  regular  hexagon  into'threeparts  such  that  the  outside  part  is 
\  the  whole  area  and  the  next  ^  of  the  whole  area. 

37.  Compute  the  sides  and  the  apothem  of  the  two  hexagons  con- 
structed in  Ex.  36  if  the  side  of  the  given  hexagon  is  12  inches. 

38.  In  the  adjoining  pattern  find  two  regular  hexagons  whose  areas 
are  in  the  ratio  1 :  4  and  show  that  this  agrees  with 
theorem,  §  348. 

39.  If  a  polygon  is  circumscribed  about  a  circle, 
show  that  the  bisectors  of  all  its  angles  meet  in  a 
point. 

40.  Given  any  polygon  circumscribed  about  a  circle. 
draw  segments  parallel  to  each  of  its  sides  and 
at  the  same  distance  from  each  side.     Show 
that  these  segments  form  a  polygon  similar  to 
the  first. 

41.  If  the  bisectors  of  all  the  angles  of  a 
polygon  meet  in  a  point  prove  that  a  circle  may 
be  inscribed  in  it  (tangent  to  all  its  sides). 

What  is  the  relation  of  the  theorems  in 
Exs.  39  and  41  ? 

42.  Given  a  polygonal  plot  of  ground  (boundary  is  a  polygon) 
such  that  a  path  of  equal  width  on  it  around  the  border  leaves  a 
polygonal  plot  similar  to  the  first.  Prove  that  a  circle  may  be  in- 
scribed in  the  polygon  which  forms  the  boundary  of  either  plot. 

43.  In  the  figure,  A  BCD  is  a  square  and  EFGHKLMNis  a  regular 
octagon. 

(a)  If  EF  =  4  inches,  find  AB.  d   l     k   C 

(6)  If  AB  =  12  inches,  find  EF. 

(c)  Ji  EF  =  a,  find  AB. 

(d)  It  AB  =  s,  find  EF. 

(e)  It  AB  =  s,  find  the  apothem. 
(/)  Jt  AB  =  s,  find  the  radius  of  the  octagon. 


REGULAR   POLYGONS  AND   CIRCLES. 


187 


44.  Find  the  ratio  between  the  areas  of  regular  octagons  inscribed 
in  and  circumscribed  about  the  same  circle. 

45.  (a)  The  apothem  of  a  regular  octagon  is  10  feet.  Find  the 
width  of  a  uniform  strip  laid  off  around  it  which  occupies  \  its  area. 

(b)  Show  how  to  construct  this  strip  geometrically  without  first 
computing  its  width.  Prove  that  the  inside  figure  is  a  regular 
octagon. 

46.  A  regular  octagon  ABCDEFGH  is  to  be  divided  into  four 
parts  of  equal  area  by  means  of  octagons  as  shown  in  the  figure. 

(a)  li  AB  =  \2  inches  find  a  side  of  each  octagon, 
(ft)  Show  how  to  construct  the  figure  without  first  computing  the 
sides. 

(c)  Show  how  to  construct  such  a  figure  if  the  four  parts  I,  II,  III, 
IV,  are  to  be  in  any  required  ratios. 

(d)  Measure  the  sides  of  the  two 
inner  parts  of  the  ceiling  pattern  and 
hence  find  the  ratio  of  their  areas. 


47.  The  middle  points,  A,B,  C,  D, 
of  alternate  sides  of  a  regular  octagon 
are  joined  as  shown  in  the  figure. 
A  H  is  perpendicular  to  A  E  and  equal 
to  it.  BG,  CK,  and  DL  are  con- 
structed in  the  same  manner. 

(a)  Prove  that  ABCD  and  HGKL 
are  squares. 

(h)  Find  the  areas  of  ABCD  and  HGKL,  if  EF=  10  inches. 

(c)  What  fraction  of  the  whole  octagon  is  occupied  by  the  square 
HGKL. 

See  the  accompanying  tile  border. 


^ 


188  PLANE  GEOMETRY. 

MEASUREMENT  OF   THE  CIRCLE. 

349.  If  in  a  circle  a  regular  polygon  is  inscribed,  its 
perimeter  may  be  measured. 

For  example,  the  perimeter 
of  a  regular  inscribed  hexa- 
gon is  6  r  if  r  is  the  radius  of 
the  circle.     See  §  334,  Ex.  10. 

If  the  number  of  sides 
of  the  inscribed  polygon  be  doubled,  the  resulting  perim- 
eter may  be  measured  or  computed  in  terms  of  r.  See 
§357. 

If  again  the  number  of  sides  be  doubled,  the  resulting 
perimeter  may  be  computed  in  terms  of  r,  and  so  on. 

In  a  similar  manner  a  regular  polygon,  say  a  hexagon, 
may  be  circumscribed  about  a  circle  and 
its  perimeter  expressed  in  terms  of  r. 

If  the  number  of  sides  of  the  circum- 
scribed polygon  be  doubled,  its  perim- 
eter may  again  be  computed,  and  so  on  as 
often  as  may  be  desired. 

350.  By  continuing  either  of  these  processes  it  is  evident 
that  the  inscribed  or  the  circumscribed  polygon  may  be 
made  to  lie  as  close  to  the  circle  as  we  please. 

351.  The  word  length  has  thus  far  been  used  in  connec- 
tion with  straight  line-segments  only.  Thus,  the  perime- 
ter of  any  polygon  is  the  sum  of  the  lengths  of  its  sides. 

352.  We  now  assume  that  a  circle  has  a  definite  length 
and  that  this  can  be  approximated  as  nearly  as  we  please 
by  taking  the  perimeters  of  the  successive  inscribed  or  cir- 
cumscribed polygons. 

The  length  of  a  circle  is  often  called  its  perimeter  or 
circumference. 


REGULAR  POLYGONS  AND   CIRCLES.  189 

It  is  evident  that  approximate  measurement  is  the  only  kind  possible 
in  the  case  of  the  circle,  since  no  straight  line  unit  of  measure,  how- 
ever small,  can  be  made  to  coincide  with  an  arc  of  a  circle. 

353.  Comparison  of  the  Lengths  of  Two  Circles.  In  each 
of  two  circles,  O  and  o',  let  regular  polygons  of  6,  12,  21, 
48,  96, 192,  etc.,  sides  be  inscribed.  Call  the  perimeters  of 
the  polygons  in  O  O,  Pg,  P^2,'>  ^24'  ^^^-i  ^i^d  those  in  O  o', 
P'g,  P'i2^  P'24,  etc.  Let  the  radii  of  the  circles  be  R  and 
r'  respectively. 

Then  by  the  theorem  of  §  347,  we  have 

4  =  ^  =  ^  =  -2.4=  etc., 

/?'         p'         pi  jy 

n         -*  6  12  2-4 

however  great  the  number  of  sides  of  the  inscribed  poly- 
gons. Show  that  the  same  relations  hold  if  polygons  are 
circumscribed  about  the  circle. 

From  these  considerations  we  are  led  to  the  following : 

354.  Preliminary  Theorem.  The  lengths  of  two 
circles  are  in  the  same  ratio  as  their  radii. 

355.  Hence,  if  C  and  &  are  the  circumferences  of  two 
circles,  R  and  r'  their  radii,  and  D  and  b'  their  diameters, 

we  have  —  =  —=  —  ;    and,  also  -  =  — :  •  (See  §  245) 

c'     r'     b'  d     n' 

Hence,  the  ratio  of  the  circumference  to  the  diameter  in 
one  circle  is  the  same  as  this  ratio  in  any  other  circle. 

356.  This  constant  ratio  is  denoted  by  the  Greek  letter  tt, 
pronounced  pi. 

The  argument  used  above  shows  that  a  theorem  like 
that  of  §  354  holds  for  every  pair  of  polygons  used  in  the 
approximation  process,  and  hence  it  is  established  for  all 
purposes  of  practical  measurement  or  computation. 


190 


PLANE  GEO  MET  BY. 


357.  Problem.     To  compute  the  approximate  value 
ofTT,  ^ 


Solution.  Suppose  a  regular  polygon  of  n  sides  is  in- 
scribed in  a  circle  whose  radius  is  r  and  let  one  of  the  sides 
AB  be  called  S^- 

We  first  obtain  in  terras  of  S^  the  length  AC^  or  S^^-*  of  a 
side  of  a  regular  polygon  oi  2n  sides  inscribed  in  the 
same  circle. 

Let  AB  be  a  side  of  the  first  polygon.  Bisect  AB  at  C 
and  draw  AC  and  BC.  Then  ^C  is  a  side  of  a  regular 
inscribed  polygon  of  2  w  sides  (Why  ?). 

Then  in  the  figure 

(82.)^=l?  =  (i^B)2+A2  =  (i5j2+^2.    (Why?)  (1) 

But  hxDE=^h(2r-h)  =  AD  xi)B==A^,  (Why?) 
or  lS'  =  a-s,y=hC2r-h).  (2) 

Solving  equation  (2)  for  A,  we  have 


2  r  ±  V4  r^  -  s,,^ 

2 


Taking  only  the  negative  sign,  since  A  <  r,  and  squaring, 
we  have, 
or, 


2r2-rV4r2-s„2_  1  s^^2^ 


^2  4.  (1  s^-)^=  2  r2  -  rV4  r^  -  8,;\ 


Hence,  from  (1),  S^„^  =  2r'^  -  rVir^  -  s,,\ 


REGULAR  POLYGONS  AND   CIRCLES,  191 


and  S2^  =  ^2r^-rV4:r^-  Sr?, 

Since,  by  §  355,  the  value  of  tt  is  the  same  for  all  circles, 
we  take  a  circle  whose  radius  is  1. 


In  this  case  -S2„  ==  ^2  _  V4  -  s„\  (3) 

If  the  first  polygon  is  a  regular  hexagon,  then  -S^g  =  1. 

Hence,      s^^  =  V2  _  VI^  =  0.51763809. 

Denoting  the  perimeter  of  a  regular  inscribed  polygon 
of  n  sides  by  P„,  we  have, 

P12  =  12(0.51763809)=  6.21165708. 

In  the  formula  (3)  let  n  =  Vl. 


Then  -^24  =  V2  _  V4  -  (0.51768809)2  =  0.26105238. 
Hence,  P^^  =  24(0,26108238)=  6.26525722. 
Computing  /S^g,  P^g,  etc.,  in  a  similar  manner,  we  have. 


s,. 

=  ^2- 

.  vri 

^ 

.51763809. 
.26105238. 
.13080626. 
.06543817. 
.03272346. 
.01636228. 
.00818126. 

-'Pu 
.-.  P,, 

•••^96 
■^'P^,2 

'•'P,S, 
•••^768 

=  6.21165708. 

Su 

=  V2- 

.  V4~ 

-  (.51763809)2  = 

P  6.26525722. 

S4S 

=  V2- 

.  V4- 

-  (.26105238)2  = 

=.  6.27870041. 

s^ 

=  V2  - 

.  VI~- 

-  (.13080626)2  = 

=  6.28206396. 

•^192 

=  V2- 

■  V4- 

-  (.06543817)2  = 

=  6.28290510. 

"^384 

=  V2  - 

■  VT^ 

-  (.03272346)2  = 

=  6.28311544. 

•^768 

=  V2- 

.  vn 

-  (.01636228)2  = 

=  6.28316941. 

358.  The  Length  of  the  Circle.  By  continuing  this  process 
it  is  found  that  the  first  five  figures  in  the  decimal  remain 
unchanged.  Hence  6.28317  is  an  approximation  to  the 
circumference  of  a  circle  whose  radius  is  1. 


192 


PLANE  GEOMETRY, 


359.  By  a  process  similar  to  the  preceding,  if  circum- 
scribed polygons  of  4,  8, 16,  etc.,  sides  are  used,  the  follow- 
ing results  are  obtained: 


Number  of  Sides 

Length  of  Each  Side 

Perimeter 

4 

2.000000 

8.000000 

8 

0.828428 

6.627418 

16 

0.397824 

6.365196 

32 

0.196984 

6.303450 

64 

0.098254 

6.288236 

128 

0.049078 

6.284448 

256 

0.024544 

6.283500 

512 

0.012272 

6.283264 

1024 

0.006136 

6.283205 

2048 

0.003068 

6.283190 

360.  Since  the  diameter  is  2,  the  ratio  of  the  circumfer- 
ence to  the  diameter  is 

64^11  =  3.14159,  or    ^•^^f^'*  =  3.14159 

according  as  the  inscribed  or  circumscribed  polygons  are 
used. 

That  is,  these  approximations  of  tt  agree  to  five  decimal 
places. 

If  so  great   accuracy  is  not  required,   we  may   use   a 

smaller  number  of  decimal  places;  such  as,  3.1416,  3.14, 

or3f 

/-» 
Since    '  =7r,  it  follows  that  C=  irD  =  27rR. 
J) 

That  is,  the  circumference  is  2  tt  times  the  radius. 


REGULAR   POLYGONS   AND   CIRCLES.  193 

361.  The  area  of  the  circle.  In  §  359  the  area  of  each 
circumscribed  polygon  is  half  the  product  of  the  perimeter 
and  the  apothem,  which  in  this  case  is  the  radius  of  the 
circle. 

That  is,  for  every  such  polygon,  -4  =  |  P  •  i?,  or  area 
equals  one  half  perimeter  times  radius. 

We  now  assume  that  a  circle  has  a  definite  area  which 
can  be  approximated  as  closely  as  we  please  by  taking 
the  areas  of  the  successive  circumscribed  polygons. 

362.  Since  the  perimeters  of  the  circumscribed  polygons 
can  be  made  to  approximate  the  length  of  the  circle  as 
nearly  as  we  please,  and  since  C  =  2  7rr  we  have, 

area  of  circle  =  |  •  2  irr  •  r  =  irr^. 

The  degree  of  accuracy  to  which  this  formula  leads 
depends  entirely  upon  the  accuracy  with  which  tt  is 
determined. 

The  old  problem  of  squaring  the  circle,  that  is,  finding  the  side  of  a 
square  whose  area  equals  that  of  a  given  circle,  involves  therefore 
determining  the  value  of  tt.  Much  time  and  labor  have  been  expended 
upon  this  in  the  hope  that  this  value  could  be  exactly  constructed  by 
means  of  the  ruler  and  compasses,  but  it  is  now  known  that  this  is 
impossible. 

363.  Since  the  area  of  a  circle  is  tt  r^,  if  we  have  two 
given  circles  whose  radii  are  r  and  r'  and  whose  diameters 
are  d  and  d' ^  then  the  ratio  of  their  areas  A  and  A'  is 

A'      7rr'^      r'^      d'^ 

Theorem.  The  areas  of  two  circles  are  in  the  same 
ratio  as  the  squares  of  their  radii,  or  of  their  diameters. 


194  PLANE  GEOMETRY. 

364.  The  area  of  a  sector  bears  the  same  ratio  to  the  area 
of  the  circle  as  the  angle  of  the  sector  does  to  the  perigon. 

E.g.  the  area  of  the  sector  whose  arc  is  a  quadrant  is  one  fourth  of 
the  area  of  the  circle,  that  is,  ^^^ — .     The  area  of  a  semicircle  is  ^^^^. 
Find  the  area  of  a  sector  whose  angle  is  60° ;  30° ;  45° ;  72°. 

365.  The  area  of  a  segment  of  a  circle  is  known  if  that  of 
the  sector  having  the  same  arc,  and  of  the  central  triangle 
on  the  same  chord,  can  be  determined. 

Show  that  the  areas  of  segments  of  a  circle  whose  arcs  are  respec- 
tively 90°  and  60°  are 

4         2        4  "^  ^  6         4  12  ^  ^ 

The  accompanying  figure  shows  a  circle  cut  into  sectors  by  a  series 
of  radii.     Each  sector  ap- 

■^  *         A         A         A         />         ,^         «         ,<         A         A         /»         A         A 

proximates  the  shape  of      i  \    i\   i  \   1  \  j  \    <  \    I  \   /  \    <  \    I  \   !  \   1  \   /  \ 
a  triangle,  whose  altitude     /    \'    \/    \/    w'    \/    \/    \/    \/    \/    \/    \/    \/    \ 
is  the  radius  of  the  circle, 

and  whose  base  is  an  arc  of  the  circle.  Since  the  sum  of  the  areas 
of  these  triangles  is  the  product  of  the  altitude  and  the  sum  of  the 
bases,  we  obtain  a  verification  of  the  theorem  that  the  area  of  a  circle 
is  one  half  the  product  of  the  circumference  and  radius. 


SUMMARY  OF  CHAPTER  V. 

1.  Make  a  list  of  the  definitions  pertaining  to  regular  polygons. 

2.  State  the  theorems  concerning  regular  polygons  inscribed  in  or 
circumscribed  about  a  circle. 

3.  State  the  theorems  involving  similar  regular  polygons. 

4.  Give  an  outline  of  the  discussion  concerning  the  value  of  tt  and 
its  use  in  approximating  the  length  and  area  of  the  circle. 

5.  What  are  some  of  the  more  important  applications  of  the  theo- 
rems in  Chapter  V?  (Return  to  this  question  after  studying  the  ap- 
plications which  follow.) 


y  ^^  ^j  ^ 

)^-©C 

)^e( 

^  r^^r 

REGULAR   POLYGONS  AND   CIRCLES.  195 

PROBLEMS  AND  APPLICATIONS  ON  CHAPTER  V. 

Note.  In  the  following  problems,  unless  otherwise  specified,  use 
the  value  rr  =  3f ,  which  differs  from  3.1416  by  about  ^^  of  one  per  cent. 

1.  The  diameter  of  a  circle  is  8  inches.  Find  the  circumference 
and  area. 

2.  The  circumference  of  a  circle  is  10  feet.  Find  its  radius  and 
area. 

3.  The  area  of  a  circle  is  24  square  inches.  Find  its  radius  and 
circumference. 

4.  Centers  of  circles  are  arranged  at  equal  distances  on  a  network 
of  lines  at  right  angles  to  each  other  as  shown  in  the 
figure.     If  r  =  I  AB,  what  part  of  the  whole  area  is 
enclosed  by  the  circles? 

Suggestion.     Consider  what  part  of  the  square 
A  BCD  lies  within  the  circles. 

5.  If  in  Ex.  4  the  radius  of  each  circle  is  3  inches, 
how  far  apart  must  the  centers  be  in  order  that  one 
half  the  area  shall  lie  within  the  circles? 

6.  If  the  radius  of  each  circle  is  r,  how  far  apart 

must  the  centers  be  located  in  order  that   -  of  the 
whole  area  may  lie  within  the  circles  ? 

7.  If  the  circles  occupy  —  of  the  area,  and  if  the  centers  are  d 
inches  apart,  find  the  radii. 

8.  A  square  is  inscribed  in  a  circle.  Find  the  ratio  between  the 
areas  of  the  square  and  the  circle. 

9.  A  square  is  circumscribed  about  a  circle.  Find  the  ratio  be- 
tween their  areas. 

Do  the  ratios  required  in  Exs.  8  and  9  depend  upon  the  radius 
of  the  circle  ? 

10.  Find  the  ratio  between  the  area  of  a  circle  and  its  regular  in- 
scribed hexagon. 

il.  Find  the  ratio  between  the  area  of  a  circle  and  its  regular  cir- 
cumscribed hexagon. 

Do  the  ratios  required  in  Exs.  10  and  11  depend  upon  the  radius  of 
the  circle  ? 


196 


PLANE  GEOMETRY. 


12.  Find  the  ratio  between  the  area  of  a  square  inscribed  in  a  circle 
and  another  circumscribed  about  a  circle  having      D  G  C 
3  times  the  radius. 

13.  Find   the  ratio  of   the  area  of  a  regular 
hexagon  inscribed  in  a  circle  to  that  of  another    jj 
circumscribed  about  a  circle  having  a  radius  \  as 
great. 

14.  J 7>CZ)  is  a  square  whose  side  is  a  and  the  ^ 
points  A^  B,  C,  D,  are  the  centers  of  the  arcs  HE,  EF,  etc. 

(a)  Find  the  area  of  the  figure  formed  by  the  arcs 
FLE,  EKH,  HNG,  GMF. 

{h)  Find  the  area  of  the  figure  formed  by  the  arcs 
ELF2indiEQF. 

Suggestion.  Find  the  difference  between  the 
areas  of  a  circle  and  its  inscribed  square. 

(c)  Find  the  area  of  the  figure  formed  by  the  arcs 
EK,  KL,  and  LE. 

15.  If  the  sides  of  the  rectangle  A  BCD  are  8  and  12  inches,  respec- 
tively, find  the  radii  of  the  circles. 

(a)  AVhat  fraction  of  the  area  of  the  rectangle 
lies  within  only  one  circle? 

(b)  Prove  that  at  each  vertex  of  a  square  two 
circles  are  tangent  to  a  diagonal  of  the  square. 

16.  Two  concentric  circles  are  such  that  one 
divides  the  area  of  the  other  into  two  equal  parts. 

(a)  Find  the  ratio  of  the  radii  of  the  circles. 

(b)  Given  the  outer  circle,  construct  the  inner  one. 

17.  Construct  three  circles  concentric  with  a  circle 
with  radius  r,  which  shall  divide  its  area  into  four  equal  parts. 

18.  Prove  that  six  circles  of  equal  radii  can  be  constructed  each 
tangent  to  two  of  the  others  and  to  a  given  circle. 

(a)  Show  that  a  circle  can  be  constructed  around 
the  six  circles  tangent  to  each  of  them. 

(b)  What  fraction  of  the  area  of  the  last  circle  is 
occupied  by  the  seven  circles  within  it? 

19.  A  OB  is  a  central  angle  of  60°.  Find  the  area 
bounded  by  the  chord  AB  and  AB  if  the  radius  of  the  circle  is  3. 


REGULAE   POLYGONS   AND   CIRCLES, 


197 


20.  ABC  is  an  equilateral  gothic  arch.    (See  page  109.)    Find  the 
area  inclosed  by  the  segment  AB  and  the  arcs  AC  and 
BC,  \iAB  =  ^  feet. 

Suggestion.  Find  the  area  of  the  sector  with  center 
A,  arc  BC,  and  radii  AB  and  AC,  aiid  add  to  this  the 
area  of  the  circle-segment  whose  chord  is  A  C. 

21.  ABC  is  an  equilateral  triangle  and  A,  B,  and  C 
are  the  centers  of  the  arcs.  Show  that  the  area  of  the 
figure  formed  by  the  arcs  is  three  times  the  area  of  one 

•  of  the  sectors  minus  twice  the  area  of  the  A  ABC. 

22.  In  the  figure  ABC  is  an  equilateral  triangle 
and  D,  E,  and  F  are  the  mid- 
dle points  of  its  sides.     Arcs 
are  constructed  as  shown. 

(a)  li  AB=6  feet,  find  the 
area  inclosed  by  CE,  EF,  FC. 

(h)  Find  the  area  inclosed 
by  .O;,  EC,  and  Cyi. 

{c)  Find  the  area  inclosed 
by  DF,  1^,  and  El). 

23.  In  the  figure  ABC  is  an  equi- 
lateral arch  and  O  0  is  constructed  on 
AB  as  a  diameter.  AH  and  BK  are 
perpendicular  to  AB. 

(a)  Construct  the  equilateral  arches 
HED  and  DFK  tangent  to  the  circle 
as  shown  in  the  figure. 

Suggestion.     0H=  OA  +  DH. 

(h)  Prove  that  the  vertices  E  and  F 
lie  on  the  circle. 

SuGGESTiox.     What  kind  of  a  triangle  is  HOKl 

In  the  following  let  AB  =  ^  feet. 

(c)  Find  the  area  bounded  by  the  arcs  DF,  FE,  and  ED. 

(d)  Find  the  area  of  the  rectangle  ABKH. 

(e)  Find  the  area  bounded  by  AH  and  the  arcs  HE  and  EA. 
(/)  Find  the  area  bounded  by  the  upper  semicircle  AB  and  the 

arcs  AC  Bind  BC. 

(g)  Find  all  the  areas  required  in  (c)  •••  (/)  if  AB  =  a. 


From  the  Uniou  Park  Church,  Chicago. 
C 


H         D         K 
From  the  First  Presbyterian 
Church,  Chicago. 


198 


PLANE  GEOMETRY. 


24.  In  the  figure,  ABCDEF  is  a  regular  hexagon.    B  is  the  center 
of  the  arc  ALC,  D  the  center  of  CHE,  and  F  the  center  of  EKA. 

li  AB  =  1Q  inches,  find  H 

(a)  The  circumference  and  area  of  the  circle, 

(b)  The  area  bounded  by  the  arc  ALC  and  the 
segments  AB  and  BC, 

(c)  The  area  bounded  by  KA,  AL  and  LK,  ^, 

(d)  The  area  bounded  by  ALC,  CHE  and  EKA. 

(e)  Find  the  areas  required  in  (a)-(d)  if  AB  =  a  inches. 

25.  ABC is  a  regular  octagon.     Arcs  are  con- 
structed with  the  vertices  as  centers  as  in  the  figure. 

(a)  If  AB=  10  inches,  find  the  area  inclosed  by 
the  whole  figure.     Also  if  AB  =  a. 

26.  ABCDEFGH  is   a  regular   octagon.      Semi- 
circles are  constructed  with  the  sides  as  diameters. 

(a)  li  AB  =  10  inches,  find  the  area  of  the  whole 
figure.     Also  \i  AB  =  a. 

(b)  Complete  the  drawing  in  the  outline  figure  to 
make  the  steel  ceiling  pattern  here  shown. 

27.  In  the  figure  A  CB,  AFD,  DEB,  and  ECF  are 
semicircles.     EF  is  tangent  to  two  semicircles. 

(a)  Prove  that  the  semicircles  AFD,  FCE,  and 
DEB  are  equal,  D  being  given  the  middle  point  of  AB. 

If  ^5  =  48  inches,  find  : 

{b)  The  area  bounded  by  AC,  7^^  and  FA, 

(c)  The  area  bounded  by  FD  and  DE,  and  the 
line-segment  FE,  A  D  B 

(d)  The  area  bounded  by  AF,  FCE,  EB,  and  the  line-segqient  AB. 

(e)  Find  the  areas  required  in  (b)-(d)  if  AB  =  a. 

28.  (a)  If  a  side  of  the  regular 
hexagon  in  the  figure  is  a,  find  the 
area  inclosed  by  the  arcs  (including 
the  area  of  the  hexagon). 

(b)  Show  that  a  circle  may  be  cir- 
cumscribed about  the  whole  figure. 

(c)  Find  the  area  inside  this  circle  and  outside  the  figure  in  (a). 


REGULAR    POLYGONS   AND   CIRCLES, 


199 


MISCELLANEOUS  PROBLEMS  AND  APPLICATIONS. 


>•<      ►^      ►-< 

X      ><      1 
►^      ^4      ►^ 

►-4      ►'^      ►-^ 


KfefefeM 


1.  Given    a  rhombus,  two  of  whose  angles   are  60°,  to   divide  it 
into    a    regular    hexagon  and    two    equilateral    tri- 
angles.    See  Ex.  7,  page  78. 

2.  What  fraction  of  the  accompanying  design  for 
tile  flooring  is  made  up  of  the  black  tiles?  Show 
how  to  construct  this  design  by  marking  off  points 
along  the  border  and  drawing  parallel  lines. 

3.  In  the  accompanying  design  for  a  parquet  border  two  strips  of 
wood  appear  to  be  intertwined. 

(a)  If  the  border  is  8  inches  wide  and  3 
feet  4  inches  long,  find  the  area  of  one  of 
these  strips,  including  the  part  which  ap- 
pears to  be  obscured  by  the  other  strip. 

(b)  If  the  figure  consists  of  squares,  find 
the  angle  at  which  the  strips  meet  the  sides.     Use  the  table  on  page  139. 

(c)  If  the  width  of  the  border  is  a  and  its  length  b,  find  the  com- 
bined area  of  these  strips. 

Compare  the  total  area  of  the  obscured  part  of  these  strips  with  the 
sum  of  the  areas  of  the  small  triangles  along  the  edge  of  the  border. 

4.  A  solid  board   fence  5  feet  in  vertical   height  running  due 
north  and  south  is  to  be  built  across  a  valley,  80  rods 
connecting  two  points  of  the  same  elevation. 
Find  the  number  of  square  feet  in  the  fence  if 
the  horizontal  distance  is  80  rods. 

5.  Are  the  data  given  in  the  preceding  problem  sufficient  to  solve 
it  if  the  fence  required  is  to  be  an  ordinary  four- 
board  fence,  each  board  6  inches  wide? 

6.  Two  circles  are  tangent  internally  at  a  point 
A.  Chords  AB  and  ^C  of  the  larger  circle  are 
drawn  meeting  the  smaller  circle  in  D  and  E 
respectively.     Prove  that  BC  and  DE  are  parallel. 

7.  Two  circles,  radii  r  and  r',  are  tangent  internally.     Find  the 
length  of  a  chord  of  the  larger  circle  tangent  to  the  smaller  if : 

(a)  The  chord  is  parallel  to  the  line  of  centers. 

(b)  The  chord  is  perpendicular  to  the  line  of  centers. 

(c)  Meets  the  larger  circle  at  the  same  point  as  the  line  of  centers. 


200 


PLANE  GEOMETRY. 


A  CBE  and  hence  dP^  =  ^. 
BE      CE 


8.  Given  a  straight  line  and  two  points  A  and  B  on  the  same 
side  of  it.  Find  a  point  C  on  the  line  such  that  the  sum  of  the  seg- 
ments A  C  and  BC  shall  be  the  least  possible. 

Solution.  In  the  figure  let  B'  be  symmetric  to  B  with 
respect  to  the  line.  Draw  AB'  meeting  the  line  in  C. 
Then  C  is  the  required  point.  For  let  C  be  any  other 
point  on  the  line.  Then  A C  +  C'B  >AC+  CB. 
The  proof  depends  upon  §  128  and  Ax.  Ill,  §  61.  (live 
it  in  full  detail. 

9.  If  in  the  figure  preceding,  A  I)  is  perpendicular 
to  the  line,  prove  that  A  A  DC 

10.  If  in  Ex.  8,  BE  =  a,  AD  =  b,  BE  =  c,  find  CD  and  CE. 

11.  Two  towns,  A  and  B,  are  10  and  6  miles  respectively  from  a  river 
and  A  is  12  miles  farther  up  the  river  than  B.     A 

pumping  station  is  to  be  built  which  shall  serve 
both  towns.  Where  must  it  be  located  so  that  the 
total  length  of  water  main  to  the  two  towns  shall 
be  the  least  possible  ? 

12.  Two  factories  are  situated  on  th^  same  side 
of  a  railway  at  different  distances  from  it.  A  spur  '  ^=-^====^' 
is  to  be  built  to  each  factory  and  these  are  to  join  the  railway  at  the 
same  point.  State  just  what  measurements  must  be  made  and  how 
to  locate  the  point  where  these  spurs  should  join  the  main  line  in  order 
to  permit  the  shortest  length  of  road  to  be  built. 

13.  Two  equal  circles  of  radius  r  intersect  so 
that  their  common  chord  is  equal  to  r.  Find  the 
area  of  the  figure  which  lies  within  both  circles. 

14.  In  the  accompanying  design  for  oak  and 
mahogany  parquet  flooring  the  large  squares  are 
6  inches  and  the  small  black  ones  2|  inches  on  a 
side.  What  fraction  of  the  whole  is  the  ma- 
hogany (the  black  squares)  ? 

15.  Construct  circles  on  the  three  sides  of  a 
right  triangle  as  diameters.     Compare  the  area 

of  the  circle  constructed  on  the  hypotenuse  with  the  sum  of  tlie  areas 
of  the  other  two.     Prove. 


K 


\ 


REGULAR  POLYGONS  AND   CIRCLES. 


201 


16.  In  the  accompanying  design  for  grill  work : 
(a)  Find  the  angles  ABC,  BAD,  and  AGC. 


(&)  If  the  radius  of  each  circle  is  r,  find  the  distance  AF. 

(c)  What  fraction  of  the  area  of  the  parallelogram  GBCE  lies 
within  one  circle  only? 

(d)  If  the  radius  of  each  circle  is  r,  find  the  distance  between  two 
horizontal  lines. 

(e)  Construct  the  whole  figure. 

Suggestions.     (1)  Find  AF  and  lay  off  points  on  AB. 

(2)  Find  Z.  BAD  and  construct  it. 

(3)  Through  the  points  of  division  on  AB  draw  lines  parallel  to 
AD. 

(4)  From  A  along  AD  lay  off  segments  equal  to  AF  and  through 
these  division  points  construct  lines  parallel  to  AB. 

(5)  Along  DC  lay  off  segments  equal  to  AF.     Connect  points  as 
shown  in  the  figure. 

(/)  From  the  construction  of  the  figure  does  it  foUow^  that 

ZDAB  =  /.EGC'i 

17.  Prove  that  the  sum  of  three  altitudes  of  a  triangle  is  less  than 
its  perimeter.     • 

18.  In  the  accompanying  design  for  grill  work,  the  arcs  are  con- 
structed from  the  vertices  of  the  equilateral  triangles  as  centers. 

(a)  Prove   that   two   arcs   are   tangent  to 
each  other  at  each  vertex  of  a  triangle. 

(b)  Find  the  area  bounded  by  the   arcs 
AB,  BC,  CD,  DA . 

(c)  Find  the  ratio  between  the  area  in  (ft) 
and  the  area  of  the  triangle  DBC. 


202 


PLANE  GEOMETRY. 


19.  The  character  of  the  accompanying  design  for  a  window  is 
obvious  from  the  figure.  Denote  the  radius  of  the  large  circle  by  R, 
of  the  semicircles  by  R' ,  and  of  the  small  circles 
by  r. 

(a)  If  i2  =:  8  feet  find  R'  and  r. 

(b)  Find  R'  and  r  in  terms  of  R. 

(c)  What  fraction  of  the  area  of  the  large  circle 
lies  within  the  four  small  circles? 

{d)  What  fraction  of  the  area  of  the  large  circle  lies  outside  the 
four  semicircles  ? 

(e)  If  it  =  10,  find  the  area  inclosed  within 
the  four  small  circles. 


20.  In  the  accompanying  design  for  a  stained 
glass  window : 

(a)  What  part  of  the  square  A'B'C'D'  lies 
within  ABCD'i 

(b)  If  A'B'  =  4:  feet,  find  the  sum  of  the  areas 
of  the  semicircles. 

(c)  Find  the  area  inclosed  by  the  line-seg- 
ments FB',  B'E  and  the  arcs  FB  and  BE,  if 
EB'  is  1^  feet. 

(d)  Find  the  areas  required  in  (b)  and  (c)  if 
A'B'  =  a. 

21.  The  accompanying  design  for  tile  flooring  consists  of  regular 
octagons  and  squares.  The  design  can  be  constructed  by  drawing 
parallel  lines  as  shown  in  the  figure. 

(a)  If  a  side  of  the  octagon  AB  is  given,  find  BC,  DE,  and  EF 
by  construction. 

Find  the  ratio  of  any  two  of  these  seg- 
ments. 

(&)  If  AB  =  a,  find  BC. 

(c)  Ji  AB  =  a  find  the  area  of  the  square 
xyzw. 

(d)  At  what  angles  do  the  oblique  lines 
meet  the  horizontal  ? 

(e)  Construct  the  figure  by  laying  off  the 
required  points  on  the  sides,  drawing  parallel  lines  in  pencil,  inking  in 
the  sides  of  the  octagons  and  erasing  the  remainder  of  the  lines. 


REGULAR   POLYGONS  AND   CIRCLES. 


203 


D 


:,x,x,x,x, 


:_x_x_x  X 


T   T   T  T  1 


T   T   X  X   : 


X    X    X    X 


r_T_T_r 


x„x  x,x 


X    X   X    X 


.X  .X    3 


^L 


<H 


22.  This  design  for  tile  flooring  is  constructed 
by  first  making  a  network  of  squares  and  then 
drawing  horizontal  lines  cutting  off  equal  tri- 
angles from  the  squares. 

(a)  At  what  angle  to  the  base  of  the  design 
are  the  oblique  lines  ? 

(h)  If  each  of  the  small  squares  is  6  inches  on 
a  side,  find  EF  and  HL. 

(c)  Find  EF  and  HL  if  the  side  of  a  small  square  is  a. 

(d)  What  fraction  of  the  whole  area  is  occupied  by  the  black 
triangles? 

23.  Five  parallel  lines  are  drawn  at  uni- 
form distances  apart,  as  shown  in  the  figure. 

(a)  If  these  lines  are  4  inches  apart,  find  the 
width  of  the  strip  from  which  the  squares 
are  made,  so  that  their  outer  vertices  shall 
just  touch  l^  and  k,  and  the  corresponding 
inner  vertices  shall  touch  Zg  and  l^. 

(b)  What  part  of  the  area  between  Z^  and 
Zg  would  be  occupied  by  a  series  of  such  squares 
arranged  as  shown  in  the  figure  ? 

24.  ABC  is  an  equilateral  triangle.  AG  and  BO 
bisect  its  base  angles.  CD  and  OE  are  drawn  parallel  to 
CA  and  CB,  respectively.     Show  that  AD  =  DE  =EB. 

25.  If  one  base  of  a  trapezoid  is  twice  the  other,  then 
each  diagonal  divides  the  other  into  two  segments  which 
are  in  the  ratio  1:2. 

26.  If  one  base  of  a  trapezoid  is  n  times  the  other,  show  that  each 
diagonal  divides  the  other  into  two  segments  which  are  in  the  ratio 
1 :  n. 

27.  Prove  that  if  an  angle  of  a  parallelogram  is  bisected,  and  the  bi- 
sector extended  to  meet  an  opposite  side,  an  isosceles  triangle  is  formed. 

Is  there  any  exception  -to  this  proposition  ?  Are  two  isosceles 
triangles  formed  in  any  case  ? 

28.  Prove  that  two  circles  cannot  bisect  each  other. 

29.  Find  the  locus  of  all  points  from  which  a  given  line-segment 
subtends  a  constant  angle. 


A    D   E   B 


204 


PLANE  GEOMETRY. 


30.  In  the  figure,  equilateral  arches  are  constructed  on  the  base  AB, 
and  on  its  subdivisions  into  halves,  fourths,  and  eighths. 

C 


\l  /^ 

x^ 

/- 

W  ArA,._y^\ 

f\ 

A 

ff^S 

A 

n 

L 

F 

D 

From  Lincoln  Cathedral,  England. 

(a)  Show  how  to  construct  the  circle  0  tangent  to  the  arcs  as  shown 
in  the  figure. 

Suggestion.  The  point  0  is  determined  by  drawing  arcs  from  A 
and  B  as  centers  with  BF  as  a  radius  (Why?). 

(6)  Show  how  to  complete  the  construction  of  the  figure. 

(c)  If  AB  =  12  feet,  find  the  radii  of  the  circles  0,  0',  0". 

(c?)  Find  these  radii  \i  AB=  s  (span  of  the  arch). 

(e)  What  part  of  the  area  of  the  arch  ABC  is  occupied  by  the  arch 
ADE'ihj  the  arch  AFS't  by  ALK'i 

(f)  The  sum  of  the  areas  of  the  seven  circles  is  what  part  of  the 
area  of  the  whole  arch? 

(g)  The  sum  of  the  areas  of  the  two  equal  circles  0'  and  O'"  is 
what  part  of  the  area  of  the  circle  O  ? 

31.  The  accompanying  church  window  design 
consists  of  the  equilateral  arch  ^-BC  and  the  six 
smaller  equal  equilateral  arches. 

(a)  li  AB  =  8  feet,  find  the  area  bounded  by  the 
arcs  MG,  GE,  ED,  DM. 

(b)  ItAB  =  S  feet,  find  the  area  bounded  by  the 
arcs^C,  CE,  ED,  DA. 

(c)  Find  the  areas  required  under  (a)  and  (b)  ii  AB 


BEGULAB  POLYGONS  AND   CIBCLES. 


205 


^K  Q 


32.  In  the  figure,  ABC  is  an  equilateral  arch. 
D,  E,  and  F,  the  middle  points  of  the  sides  of  the 
triangle  ABC,  are  centers  of  the  arcs  AE,  KL, 
BF,  and  SR ;  CF  and  MR ;  EC  and  LM  re- 
spectively. 

(a)  Prove  that  the  arc  with  center  D  and  radius 
DA  passes  through  the  point  E. 

(b)  Prove  that  arcs  with  centers  D  and  F,  and  tangent  to  the  seg- 
ment A  C,  meet  on  the  segment  BE. 

(c)  If  AB  =  a,  find  KS. 

(d)  Can  we  find  the  area  bounded  by  the  segment  ^  J5  and  the  arcs 
BF,  FC,  CE,  and  EA  when  ^^  is  given  ?  If  so,  find  this  area  when 
AB  =  a. 

(e)  Can  we  find  the  area  bounded  by  KS  and  the  arcs  SR,  RM, 
ML,  LK,  when  AB  is  given  ?    If  so,  find  this  area  when  AB  =  6  feet. 

33.  Prove  that  the  altitude  of  an  equilateral  triangle  is  three  times 
the  radius  of  its  inscribed  circle. 

34.  The  accompanying  grill  design  is 
based  on  a  network  of  congruent  equilateral 
triangles.  Arcs  are  constructed  with  ver- 
tices of  the  triangles  as  centers. 

(a)  If  AB  =  Q  inches,  find  the  area 
bounded  by  CQ,  QP,  PT,  fs,  SR,  RC. 

(h)  Has  the  figure  consisting  of  these  arcs  a  center  of  symmetry  ? 
How  many  axes  of  symmetry  has  it? 

(c)  Find  the  area  required  under  (a)  \i  AB  =  a. 

(d)  liAB  =  4:  inches,  find  the  area  bounded  by  CD,  DE,  EF,  FG, 
GH,  HK,  etc. 

(e)  Has  the  figure  consisting  of  these  arcs  a  center  of  symmetry? 
How  many  axes  of  symmetry  has  it? 

(/)    Find  the  area  required  under  (d)  if  AB  =  a. 


35.   Two  circles  intersect  in  the  points  A  and  B 
line  is  drawn,  meeting  the  two  circles 
in  C  and  D  respectively,  and  through     9^ZZ^A 
B  one  is  drawn  meeting  the  circles 
in  E  and  F  respectively.     Prove  that 
CE  and  DF  are  parallel. 


Through  A  a 


206 


PLANE  GEOMETRY. 


36.  Prove  that  if  the  points  D  and  F  coincide  in  the 
preceding  example  the  tangent  at  D  is  parallel  to  CE. 

37.  Two  circles  are  tangent  internally  at  J..  Prove 
that  all  chords  of  the  larger  circle  through  A  are  di- 
vided proportionally  by  the  smaller  circle. 

38.  Chords  are  draw^n  through  a  fixed  point  on  a  circle.     Find  the 
locus  of  points  which  divide  them  into  a  fixed  ratio. 

39.  Squares  are  inscribed  in  a  circle,  a  semicircle,  and  a  quadrant 
of  the  same  circle.     Compare  their  areas. 


40.  In  a  given  circle  two  diameters  are  drawn  at  right  angles  to 
each  other.  On  the  radii  thus  formed  as  diameters  semicircles  are 
constructed.     Show  that  the  four  figures  thus  formed  are  congruent. 

41.  Let  C  be  any  point  on  the  diameter  AB  oi  2i,  circle. 

(a)  Compare  the  length  of  the  arc  ADB  with  the  sum  of  the 
lengths  of  the  arcs  AEC  and  CFB. 

(b)  Show  that  ii  AB  =  ^  CB,  then  the  area  inclosed  by  the  arcs 
BFC,  CEA,  ADB,  is  one  third  the  area  of  the  circle. 

(c)  Show  that  if  AB  =  m-  CB,  then  the  area  inclosed  by  these 
arcs  is  one  mth  of  the  area  of  the  circle. 

42.  By  means  of  arcs  constructed  as  shown  in  the  third  figure 
divide  the  area  of  a  circle  into  any  given  number  c'^ 

of  equal  parts.     Make  the  construction.  \"v^ 

43.  Two  sides  AB  and  BC  oi  a,  triangle  are  ex- 
tended their  own  lengths  to  B'  and  C  respectively. 
Compare  the  areas  of  the  triangles  ABC  and 
BB'C. 

44.  The  three  sides  of  a  triangle  ABC  are  ex- 
tended to  A',  B',  C  as  shown  in  the  figure.  Com- 
pare the  areas  of  the  triangles  ABC  and  A'B'C : 

(a)  HBB'  =  AB,  CC  =  BC,  and  A  A'  =  CA  ; 
(h)iiBB'  =  l.AB,CC'  =  m.BC,aiidAA'  =  rk-CA.    a' 


JO"- 


B,-'' 


CHAPTER   YI. 


VARIABLE    G-BOMETRIO    MAGNITUDES. 


GRAPHIC  REPRESENTATION. 


366.  It  is  often  useful  to  think  of  a  geometric  figure  as 
continuously  varying  in  size  and  shape. 

E.g.  if  a  rectangle  has  a  fixed  base,  say  10  inches  long,  but  an  al- 
titude which  varies  continuously  from  3  inches  to  5  inches,  then  the 
area  varies  continuously  from  3  .  10  =  30  to  5  •  10  =  50  square  inches. 

We  may  even  think  of  the  altitude  as  starting  at  zero  inches  and 
increasing  continuously,  in  which  case  the  area  starts  at  zero  and  in- 
creases continuously. 

From  this  point  of  view  many  theorems  may  be  repre- 
sented graphically.  The  graph  has  the  advantage  of  ex- 
hibiting the  theorem  for  all  cases  at  once. 

For  a  description  of  graphic  representation  see  Chapter  V  of  the 
authors'  High  School  Algebra,  Elementary  Course. 

367.  If  in  the  figure  A  C  II  BD  and  OA  and  OB  are  commen- 
surable., and  if  —  =  ^^ ,  then  O,  C,  and  D  lie  in  a  straight 
,.  OB       BD 

line. 

For  suppose  D  not  in  a  line  with  DC.     Pro- 
duce DC  and  BD  to  meet  at  K. 
OA  ^AC 
OB      BK 
OA^AC 
OB      BD 


Then 

But  by  hypothesis 


BD  =  BK. 


(Why?) 


Hence, 

Therefore  D  coincides  with  K,  and  0,  C  and  D  are  in  a  straight 
line. 

207 


208 


PLANE  GEOMETRY. 


368.  Theorem.    The  areas  of  two  rectangles  having 
equal  bases  are  in  the  same  ratio  as  their  altitudes. 

Graphic  Representation.     For  rectangles  with  commensurable  bases 
and  altitudes  we  have 

Area  =  base  x  altitude. 


Consider  rectangles  each  with  a  base  equal  to  b,  altitudes 
etc.,  each  commensurable  with  b,  and  areas  A-^,  A 2,  A^,  etc. 


Then, 


bh. 


A, 


',  etc. 


(§303) 


(1) 


We  exhibit  graphically  the  special  case  where  6  =  10.  Let  one 
horizontal  space  represent  one  unit  of  altitude  and  one  vertical  space 
ten  units  of  area. 

Thus,  the  point  P^  has  the  ordinate  ^  =  10  vertical  units  (repre- 
senting 100  units  of  area)  and  the  abscissa  ^3  =  1^  horizontal  units. 

Similarly  locate  Pi  and  Pg  whose  abscissas  are  hi  and  A3  and  ordi- 
nates  Ai  and  A2. 

Using  equation  (1)  and  §  367,  show  that  0,  P^,  P2,  Pz  He  in  the 
same  straight  line. 


n 

A 

3 

^8 

"" 

>-! 

/ 

/ 

t» 

I 

>/ 

5 

1 

■ 

,._ 

/ 

7 

— 

< 

h 

=  1  (V 

'f. 

^/ 

1  * 

/ 

/ 

1 

h 

2= 

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S 

VARIABLE  GEOMETRIC  MAGNITUDES.  209 

If  we  suppose  that  while  the  base  of  the  rectangle  re- 
mains fixed,  the  altitude  varies  continuously  through  all 
values  from  Ag  =  10  to  Ag  =  20  =  2  x  10,  then  it  must  take 
among  other  values  the  value  10  V2. 

Using  10  V2  as  an  abscissa,  the  question  is,  whether 
CD  is  the  area  ordinate  corresponding  to  it. 

This  area  ordinate  is  not  less  than  CD  and  commen- 
surable with  the  base,  for  in  that  case  the  altitude  would 
be  less  than  10  V2,  and  for  the  same  reason  it  is  not 
greater  than  CD  and  commensurable  with  the  base. 

But  we  can  find  line-segments  less  than  10  V2  or  greater 
than  10  V2  and  as  near  to  10  V2  as  we  please. 

Hence  we  conclude  that  the  area  ordinate  for  the 
rectangle  whose  altitude  is  10  V2  is  CD,  that  is,  the  point  c 
lies  in  the  line  OP^. 

In  like  manner,  the  point  determined  by  any  other  ab- 
scissa incommensurable  with  the  base  is  shown  to  lie  on 
the  line  OP^^. 

Since  the  abscissa  and  ordinate  of  any  point  on  OP  are 
equal,  we  have  for  any  altitude, 

369.   The  preceding  theorem  may  also  be  stated : 
The  area  of  a  rectangle  ivith   a  fixed  base  varies 
directly  as  its  altitude. 

This  means  that  if  A  and  h  are  the  varying  area  and  altitude  re- 
spectively, and  if  ^1  and  hi  are  the  area  and  altitude  at  any  given 
instant,  then 

-— -  =  —  or  ^  =  — ^  '  h  or  A  =  kh,  where  k  is  the  Jixed  ratio  —^' 
A\      h,\  hi  hi 

The  graph  representing  the  relations  of  two  variables 
when  one  varies  directly  as  the  other  is  always  a  straight  line. 


210  PLANE    GEOMETBY, 

370.  EXERCISES. 

1.  Make  a  graph  to  show  that  the  area  of  two  rectangles  having 
equal  altitudes  are  in  the  same  ratio  as  their  bases. 

2.  Show  by  a  graph  that  the  area  of  a  triangle  having  a  fixed  alti- 
tude varies  as  the  base,  and  having  a  fixed  base  varies  as  the  altitude. 

3.  Represent  graphically  the  relation  between  two  line-segments 
both  of  which  begin  at  zero,  and  one  of  which  increases  three  times 
as  fast  as  the  other.     Five  times  as  fast.     One  half  as  fast. 

4.  If  areas  be  represented  by  the  length  of  a  line-segment  as  in 
§  368,  which  question  in  Ex.  3  applies  to  the  altitude  and  area  of  a 
parallelogram  having  a  fixed  base  and  varying  altitude?  Which 
applies  to  a  triangle  having  a  fixed  base  and  varying  altitude? 

371.  Theorem.  The  area  of  a  rectangle  is  equal  to 
the  product  of  its  base  and  altitude. 

Proof :  Using  the  graphic  representations  of  §§  368, 
370,  Ex.  1,  we  have  for  all  cases, 

^  =  ?  and  ^=1. 


a 

^1 

1 

^, 

1 

Az 

b 

^1        1  ^^2        1 

Hence     ^  =  ^  x  ^  =  ?  x -  =  a5. 
A^     A^     A,^     1      1 

But  A^  is  the  unit  of  area.     Hence 
ab  represents  the  numerical  measure  of  " 

^3  by  the  area  unit. 

That  is,  Area  =  base  x  altitude. 

372.  Problem.  Make  a  graphic  representation  of 
the  theorem  :  The  perimeters  of  similar  polygons  are 
in  the  same  ratio  as  anyjivo  corresponding  sides. 

Solution.  First  consider  the  special  case  of  equilat- 
eral triangles.  On  the  horizontal  axis  lay  off  the  lengths 
of  one  side  of  several  such  triangles,  and  on  the  vertical 


VARIABLE  GEOMETRIC  MAGNITUDES. 


211 


axis  lay  off  the  lengths  of  the  cor- 
responding perimeters.  Show  that 
the  points  so  obtained  lie  in  a  straight 
line. 

373.  EXERCISES. 

1.  In  the  manner  above  graph  the  rela- 
tion between  the  perimeters  and  sides  of 
squares.  Of  regular  pentagons.  Of  regu- 
lar hexagons.     Of  rhombuses. 

2.  If  a  side  of  a  given  regular  polygon  is 
a  and  its  perimeter  p,  graph  the  relation  of 
the  perimeters  and  corresponding  sides  of 
polygons  similar  to  the  given  polygon. 


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374.  Theorem,  If  a  line  is  parallel  to  one  side  of  a 
triangle  and  cuts  the  other  tioo  sides,  then  it  divides  these 
two  sides  in  the  same  ratio.  c 

Graphic  representation : 

Lay  off  CA  along  the  hori- 
zontal axis  and  CB  along  the 
vertical  axis,  thus  locating  the  A 
point  Pj. 

In  like   manner  find  P^  with  the 
coordinates  CD  and  CE.    If  CD  and  CA 


are  commensurable,  we  know  that 
CD  _CE 
CA  ~  Cb' 
Hence  o,  Pg  and  Pj  are  collinear. 
If  CD'  and  CA   are  incommensur- 
able, show,  as  in  §  368,  that  the  corresponding  point  P 
lies  on  the  line  OP^^  and  hence,  as  in  that  case,  also 
CD^  _  CE^ 
CA  ~~  cb' 


CA 


212 


PLANE  GEOMETBY. 


375.  Pkoblem.  To  represent  graphically  the  rela- 
tion hetiveen  the  area  and  side  of  a  square  as  the  side 
varies  conthiuously. 


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12345678  Axir  for  Sides 

Solution.  On  the  horizontal  axis  lay  off  the  segments 
equal  to  various  values  of  the  side  s,  and  on  the  vertical 
axis  lay  off  segments  equal  to  the  corresponding  areas  A, 

(1)  If  one  horizontal  space  represents  one  unit  of  length 
of  side,  and  one  vertical  space  one  unit  of  area,  then  the 
points  Pj,  Pg,  P3,  etc.,  are  found  to  lie  on  the  steep  curve. 


VARIABLE  GEOMETRIC  MAGNITUDES.  213 

(2)  If  five  horizontal  spaces  are  taken  for  one  unit  of 
length  of  side  and  one  vertical  space  for  one  unit  of  area, 
then  the  points  P^\  P^^  Pg',  etc.,  are  found  and  the  less 
steep  curve  is  the  result. 

The  student  should  locate  many  more  points  between  those  here 
shown  and  see  that  a  smooth  curve  can  be  drawn  through  them  all  in 
each  case. 

The  graph  of  the  relation  between  two  variables,  one  of  which 
varies  as  the  square  of  the  other,  is  always  similar  to  the  one  here 
given. 

376.  The  area  of  a  square  is  said  to  vary  as  the  square 
of  one  of  its  sides,  that  is,  A  =  s^. 

For  example,  the  theorem  :  The  areas  of  two  similar  polygons  are  in 
the  same  ratio  as  the  squares  of  any  two  corresponding  sides,  means  that 
if  a  given  side  of  a  polygon  is  made  to  vary  continuously  while  the 
polygon  remains  similar  to  itself,  the  area  of  the  polygon  varies  con- 
tinuously as  the  square  of  the  side. 

377.  EXERCISES. 

1.  From  the  last  graph  find  approximately  the  areas  of  squares 
whose  sides  are  3.4  ;  5.25;  6.35. 

2.  Find  approximately  from  the  graph  the  side  of  a  square  whose 
^  area  is  28  square  units ;  21  square  units ;  41.5  square  units. 

3.  Construct  a  graph  showing  the  relation  between  the  areas  and 
sides  of  equilateral  triangles. 

4.  Given  a  polygon  with  area  A  and  a  side  a.  Construct  a  graph 
showing  the  relation  between  the  areas  and  the  sides  corresponding 
to  a  in  polygons  similar  to  the  one  given. 

5.  From  the  graph  constructed  in  Ex.  3,  find  the  area  of  an  equi- 
lateral triangle  whose  sides  are  4.  Also  of  one  whose  sides  are  6. 
Compute  these  areas  and  compare  results. 

6.  Construct  a  graph  showing  the  relation  between  a  side  and  the 
area  of  a  regular  hexagon.  By  means  of  it  find  the  area  of  a  regular 
hexagon  whose  sides  are  6.     Compare  with  the  computed  area. 


214 


PLANE  GEOMETBY. 


378.  Problem.  To  construct  a  graph  showing  the 
relation  between  the  radius  and  the  circumference  of  a 
circle,  and  between  the  radius  and  area  of  a  circle^  as 
the  radius  varies  co7itinuously . 

Solution.  Taking  ten  horizontal  spaces  to  represent 
one  unit  of  length  of  radius,  and  one  vertical  space  for  one 
unit  of  circumference  in  one  graph  and  one  unit  of  area 
in  the  other,  we  find  the  results  as  shown  in  the  figure. 


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^                     ^^          ^^ 

^^          ^^ 

^^          ^^ 

^^          ^^ 

^^       ^^^ 

Axis  for  Radii 


VABIABLE  GEOMETRIC  MAGNITUDES.  215 

379.  EXERCISES. 

1.  From  the  graph  find  approximately  the  circumference  of  a 
circle  whose  radius  is  2.7,  also  of  one  whose  radius  is  3.4. 

2.  Find  the  radius  of  a  circle  whose  circumference  is  17,  also  of 
one  whose  circumference  is  23. 

3.  Find  the  areas  of  circles  whose  radii  are  1.9,  2.8,  3.6. 

4.  Find  the  radii  of  circles  whose  areas  are  13.5,  25.5,  37,  45. 

5.  How  does  the  circumference  of  a  circle  vary  with  respect  to  tHe 
radius? 

6.  How  does  the  area  of  a  circle  vary  with  respect  to  the  radius  ? 

7.  Find  the  radius  of  that  circle  whose  area  in  square  units  equals 
its  circumference  in  linear  units. 

DEPENDENCE  OF  VARIABLES. 

380.  In  the  preceding  pages  we  have  considered  certain 
areas  or  perimeters  of  polygons  as  varying  through  a 
series  of  values.  For  example,  if  a  rectangle  has  a  fixed 
base  and  varying  altitude,  then  the  area  also  varies  de- 
pending on  the  altitudes.  The  fixed  base  is  called  a  con- 
stant, while  the  altitude  and  area  are  called  variables. 

The  altitude  which  we  think  of  as  varying  at  our  pleas- 
ure is  called  the  independent  variable,  while  the  area, 
being  dependent  upon  the  altitude,  is  called  the  dependent 
variable. 

381.  The  dependent  variable  is  sometimes  called  a  func- 
tion of  the  independent  variable,  meaning  that  the  two 
are  connected  by  a  definite  relation  such  that  for  any 
definite  value  of  the  independent  variable,  the  dependent 
variable  also  has  a  definite  value. 

Thus,  in  A  =  s'^  (§  376),  J  is  a  function  of  s,  since  giving  s  any 
definite  value  also  assigns  a  definite  value  to  A . 

Similarly,  C  is  a  function  of  r  in  C  =  2  irr,  and  ^  is  a  function  of 
rin  A  =  irr^. 


216  PLANE  GEOMETRY. 

382.  EXERCISES. 

Justify  each  of  the  followmg  statements,  remembering  that  a 
variable  y  varies  directly  as  another  variable  x  ii  y  =  kx,  and  directly 
as  the  square  oi  x  ii  y  =  kx\  where  k  is  some  constant.  Find  k  in  each 
case. 

1.  The  area  of  a  rectangle  with  constant  base  varies  directly  as 
its  altitude.     Find  the  value  of  k. 

Suggestion.     By  §  368,  4-  =  T-   ^^  A=^.h.    Hence  A  =  kh. 
Ai      ti-y  hy 

In  this  case  k  =  — ^=b,  the  constant  base.     See  also  §  369. 
^i 

2.  The  area  of  a  square  varies  directly  as  the  square  of  its  side. 

Suggestion.     Since  A=s^    and    A^  =  s^^,  we   have  —  =  —  or 

4  A  ^'     ''' 

A=^'  s^     That  is,  A  =  ks^  where  k=^=l.     See  §  376. 

3.  An  angle  inscribed  in  a  circle  varies  directly  as  the  intercepted 
arc.     Show  that  in  this  case  k  =  I. 

4.  A  central  angle  in  a  circle  varies  as  the  intercepted  arc.  Show 
that  in  this  case  k  =  1. 

5.  In  the  figure  of  §  374,  show  that  DE  varies  directly  as  CD  if 
DE  moves,  remaining  parallel  to  AB. 

6.  An  angle  formed  by  two  chords  intersecting  within  a  circle 
varies  directly  as  the  sum  of  the  two  arcs  intercepted  by  the  angle 
and  its  vertical  angle. 

7.  An  angle  formed  by  two  secants  intersecting  outside  a  circle 
varies  directly  as  the  difference  of  the  two  intercepted  arcs. 

8.  If  a  polygon  varies  so  as  to  remain  similar  to  a  fixed  polygon, 
then  its  perimeter  varies  directly  as  any  one  of  its  sides.     Show  that 

p 
k  =  —I,   where  P,  and  s,    are   the  perimeter  and  side  of  the   fixed 

h 
polygon. 

9.  In  the  preceding,  the  area  of  the  polygon  varies  directly  as  the 
square  of  any  one  of  its  sides. 

10.  The  circumference  of  a  circle  varies  directly  as  the  radius, 
and  its  area  varies  directly  as  the  square  of  its  radius. 


VARIABLE  GEOMETRIC  MAGNITUDES.  217 

11.  Plot  y  =  kx  for  Jc  =  1,  h  3,  4. 

12.  Plot  y  =  kx^  for  yfc  =  1,  i  3,  4. 

Notice  that  the  graphs  in  Ex.  11  are  all  straight  lines,  while  those 
in  Ex.  12  are  curves  which  rise  more  and  more  rapidly  as  the  inde- 
pendent variable  increases.     See  also  §  378. 

LIMIT  OP  A  VARIABLE. 

383.  If  a  regular  polygon  (§  357)  is  inscribed  in  a  circle 
of  fixed  radius,  and  if  the  number  of  sides  of  the  polygon 
be  continually  increased,  for  instance  by  repeatedly 
doubling  the  number,  then  the  apothem,  perimeter,  and 
area  are  all  variables  depending  upon  the  number  of  sides. 
That  is,  each  of  these  is  a  function  of  the  number  of  sides. 

Now  the  greater  the  number  of  sides  the  more  nearly 
does  the  apothem  equal  the  radius  in  length.  Indeed,  it 
is  evident  that  the  difference  between  the  apothem  and 
the  radius  will  ultimately  become  less  than  any  fixed 
number,  however  small.  Hence  we  say  that  the  apothem 
approaches  the  radius  as  a  limit  as  the  number  of  sides 
increases  indefinitely. 

384.  Similarly  by  §  352  the  perimeters  of  the  polygons 
considered  in  the  preceding  paragraph  may  be  made  as 
nearly  equal  to  the  circumference  as  we  please  by  making 
the  number  of  sides  sufficiently  great.    - 

Hence  we  define  the  circumference  of  a  circle  as  the 
limit  of  the  perimeter  of  a  regular  inscribed  polygon  as 
the  number  of  sides  increases  indefinitely. 

It  also  follows  from  §  353  that  the  circumference  of  a 
circle  may  be  defined  as  the  limit  of  the  perimeter  of  a 
circumscribed  polygon  as  the  number  of  sides  is  increased. 

Likewise  we  may  define  the  area  of  a  circle  as  the  limit 
of  the  area  of  the  inscribed  or  the  circumscribed  polygon 
as  the  number  of  sides  increases  indefinitely.     See  §  361. 


218  PLANE  GEOMETRY. 

385.  The  notion  of  a  limit  may  be  used  to  define  the 
length  of  a  line-segment  which  is  incommensurable  with 
a  given  unit  segment. 

Thus,  the  diagonal  c?  of  a  square  whose  side  is  unity  \s  d  =  V2. 
Hence  d  may  be  defined  as  the  limit  of  the  variable  line-segment  whose 
successive  lengths  are  1,  1.4,  1.41,  1.414- •-.     See  §§  234,  240. 

In  like  manner,  the  length  of  any  line-segment,  whether 
commensurable  or  incommensurable  with  the  unit  seg- 
ment, may  be  defined  in  terms  of  a  limit. 

Thus,  if  a  variable  segment  is  increased  by  successively  adding  to  it 
one  half  the  length  previously  added,  then  the  segment  will  approach 
a  limit.  If  the  initial  length  is  1,  then  the  successive  additions  are  I, 
I,  \,  j\,  -^^,  etc.,  and  the  successive  lengths  are  1,  1|,  If,  1|,  li|,  lf|, 
etc.     Evidently  this  segment  approaches  the  limit  2. 

Hence  2  may  be  defined  as  the  limit  of  the  variable  segment, 
whose  successive  lengths  are  1,  1^,  1|,  1|,  etc.,  as  the  number  of  suc- 
cessive additions  is  increased  indefinitely. 

386.  A  useful  definition  of  a  tangent  to  a  circle,  or  to 
any  other  smooth  curve,  may  be  given  in  terms  of  a  limit. 

Let  a  secant  cut  the  curve  in  a   fixed   point  A  and  a 
variable  point  B,  and  let  the  point  B 
move  along   the  curve  and  approach 
coincidence  with  A,  thus  making  the 
secant  continually  vary  its  direction. 

Then  the  tangent  is  defined  to  be 
the  limiting  position  of  the  secant  as 
B  approaches  A  indefinitely. 

This  definition  of  a  tangent  is-  used  in  all  higher  mathematical 
work.     It  includes  the  definition  given  in  the  case  of  the  circle  in  §  183. 

387.  The  functional  relation  between  variables  and  the 
idea  of  a  limit  as  illustrated  above  are  two  of  the  most  im- 
portant concepts  in  all  mathematics.  The  whole  subject  is 
much  too  difficult  for  rigorous  consideration  in  this  course. 


VABIABLE  GEOMETRIC  MAGNITUDES.  219 

388.  EXERCISES. 

1.  Find  the  limit  of  a  variable  line-segment  whose  initial  length 
is  6  inches,  and  which  varies  by  successive  additions  each  equal  to 
one  half  the  preceding. 

2|.  Find  the  limit  of  a  variable  line-segment  whose  initial  length 
is  1  and  whose  successive  additions  are  .3,  .03,  .003,  etc. 

3.  Construct  a  right  triangle  whose  sides  are  1  and  2.  By  approxi- 
mating a  square  root,  find  five  successive  lengths  of  a  segment  which 
approaches  the  length  of  the  hypotenuse  as  a  limit. 

4.  If  one  tangent  to  a  circle  is  fixed,  and  another  is  made  to  move 
so  that  their  intersection  point  approaches  the  circle,  what  is  the 
limiting  position  of  the  moving  tangent  ?  What  is  the  limit  of  the 
measure  of  the  angle  formed  by  the  tangents  ? 

5.  The  arc  ^^  of  74°  is  the  greater  of  the  two  arcs  intercepted  be- 
tween two  secants  meeting  at  C  outside  the  circle.  The  points  A  and  B 
remain  fixed  while  C  moves  up  to  the  circle.  What  is  the  limit  of  the 
angle  formed  by  the  secants  ?    The  limit  of  the  measure  of  their  angle  ? 

6.  If  in  the  preceding  the  secants  meet  within  the  circle,  what  is 
the  limit  of  their  angle  and  also  of  the  measure  of  this  angle  ? 

7.  If  in  Ex.  5  one  secant  and  the  intersection  point  remain  fixed, 
while  the  other  secant  approaches  the  limiting  position  of  a  tangent 
at  the  point  A,  find  the  limit  of  the  measure  of  the  included  angle. 

8.  If  in  Ex.  6  one  secant  and  the  point  of  intersection  remain 
fixed  while  the  other  secant  swings  so  as  to  make  the  included  angle 
approach  a  straight  angle,  find  the  limit  of  the  measure  of  the  angle. 

9.  If  in  Exs.  5  and  6  the  moving  point  crosses  the  circle,  state  the 
theorem  on  the  measurement  of  the  angle  in  question  so  as  to  apply 
equally  well  whether  the  point  is  inside  or  outside  the  circle. 

10.  A  fixed  segment  ^  J5  is  divided  into  equal  parts,  and  equilateral 
triangles  are  constructed  on  each 
part  as  a  base,  as  A  EEC,  CGD, 
DHA.  Then  each  base  is  divided 
into  equal  parts  and  equilateral  tri- 
angles are  constructed  on  these  parts 
as  bases.  What  is  the  limit  of  the 
sum  of  the  perimeters  of  these  triangles  as  the  number  of  them  is 
increased  indefinitely  ?    What  is  the  limit  of  the  sum  of  their  areas  ? 


CHAPTER  VII. 

REVIEW  AND  FURTHER  APPLICATIONS. 

ON  DEFINITIONS  AND  PROOFS. 

389.  A  definition  is  a  statement  that  a  certain  word  or 
phrase  is  to  be  used  in  place  of  a  more  complicated 
expression. 

Thus  the  word  "triangle"  is  used  instead  of  "the  figure  formed 
by  three  segments  connecting  three  non-collinear  points." 

In  geometry  we  may  distinguish  two  classes  of  words : 

(a)  Technical  words  representing  geometric  concepts 
such  as  line,  plane,  polygon,  circle,  etc. 

(5)  Words  of  ordinary  speech  not  included  in  the  first 
class. 

The  meaning  of  words  in  the  second  class  is  taken  for 
granted  without  any  definition  whatever.  It  is  not 
possible  to  define  everi/  word  of  the  first  class,  for  every 
definition  brings  in  new  technical  terms  which  in  turn 
require  definition. 

Thus,  one  of  the  many  definitions  of  "point"  is  "that  which 
separates  one  part  of  a  line  from  the  adjoining  part." 

In  this  definition  the  technical  terms  "  separate,"  "  line,"  "  adjoin- 
ing "  are  used.  If  we  try  to  define  these,  still  other  terms  are  brought 
in  which  need  to  be  defined,  and  so  on.  The  only  escape  is  defining 
in  a  circle,  which  is  not  permitted  in  a  logical  science. 

Since  it  is  thus  evident  that  some  terms  must  be  used 
without  being  defined,  it  is  best  to  state  which  ones  are 
left  undefined. 

220 


REVIEW  AND  FUBTHER  APPLICATIONS.  221 

390.  In  this  book  point,  straight  line,  or  simply  line, 
plane,  size  and  shape  of  figures  are  not  defined.  Nor  do 
we  define  the  expression,  a  point  is  between  two  other 
points,  or  a  point  lies  on  a  segment. 

Descriptions  of  some  of  these  terms  which  do  not,  how- 
ever, constitute  definitions  in  the  logical  sense  are  given 
in  §§  1-5. 

Other  technical  terms  are  defined  by  means  of  these 
simple  undefined  words  with  which  we  start. 

Thus,  "a  segment  is  that  part  of  a  line  which  lies  hetioeen  two  of 
its  points  "  is  a  definition  of  segment  in  terms  of  the  three  undefined 
words,  point,  line,  and  between. 

391.  A  geometric  proposition  consists  of  an  affirmation 
that  a  geometric  figure  has  some  property  not  explicitly 
specified  in  its  definition. 

A  proposition  is  said  to  be  proved  if  it  is  shown  to  fol- 
low from  '  other  propositions  which  are  admitted  to  be 
true.  Hence  every  geometric  proposition  demands  for  its 
proof  certain  other  propositions. 

392.  It  is  obvious,  therefore,  that  certain  propositions 
must  be  admitted  without  proof.  Such  unproved  propo- 
sitions are  called  axioms. 

In  order  that  a  set  of  axioms  for  geometry  shall  be 
complete  it  must  be  possible  to  prove  that  every  theorem 
of  geometry  follows  from  them.  The  set  of  axioms  used 
in  this  text  is  not  complete. 

For  instance,  it  is  assumed  without  formal  statement  that  if  ^,  B, 
C,  are  three  points  on  a  segment  we  cannot  have  at  the  same  time  B 
between  A  and  C  and  C  between  A  and  B ;  that  the  diagonals  of  a 
convex  quadrilateral  intersect  each  other ;  that  a  ray  drawn  from  the 
vertex  of  an  angle  and  included  between  its  sides  intersects  every 
segment  determined  by  two  points,  one  on  each  side  of  the  angle. 
In  like  manner  many  other  tacit  assumptions  are  made. 


222  PLANE  GEOMETRY. 

393.  In  a  complete  logical  treatment  every  undefined 
term  must  occur  in  one  or  more  axioms,  since  all  knowl- 
edge of  this  term  in  a  logical  sense  comes  from  the  axioms 
in  which  it  is  found.  In  this  text  not  every  undefined 
term  occurs  in  an  axiom,  for  instance,  the  word  hetiveen. 

The  axioms  are,  of  course,  based  on  our  space  intuition^ 
or  on  our  experience  with  the  space  in  which  we  live.  It 
is  interesting  to  notice,  however,  that  the  axioms  tran- 
scend that  experience  both  as  to  exactness  and  extent.  For 
instance,  we  have  had  no  experience  with  endless  lines, 
and  hence  we  cannot  know  directly  from  experience 
whether  or  not  there  are  complete  lines  which  have  no 
point  in  common.     See  §§  89,  96. 

394.  Proofs  are  of  two  kinds,  direct  and  indirect.  A 
direct  proof  starts  with  the  hypothesis  and  leads  step  by 
step  to  the  conclusion. 

An  indirect  proof  starts*  with  the  hypothesis  and  with 
the  assumption  that  the  conclusion  does  not  hold,  and 
shows  that  this  leads  to  a  contradiction  with  some  known 
proposition.  Or  it  starts  simply  with  the  assumption 
that  the  conclusion  does  not  hold  and  shows  that  this 
leads  to  a  contradiction  with  the  hypothesis.  This  kind 
of  proof  is  based  upon  the  logical  assumption  that  a 
proposition  must  either  be  true  or  not  true.  The  proof 
consists  in  showing  that  if  the  proposition  were  not  true, 
impossible  consequences  would  follow.  Hence  the  only 
remaining  possibility  is  that  it  must  be  true. 

395.  Every  proposition  in  geometry  refers  to  some 
figure.  See  §  12.  The  essential  characteristic  of  a  figure 
is  its  description  in  words  and  not  the  drawing  that  repre- 
sents it.  Each  drawing  represents  just  one  figure  from 
a  class  of  figures  defined  by  the  description.     Thus  we  say 


BEVIEW  AND  FURTHER   APPLICATIONS.  223 

"let  ABCD  be  a  convex  quadrilateral,"  and  we  construct 
a  particular  quadrilateral.  We  must  then  take  care  that 
all  we  say  about  it  applies  to  ani/  figure  whatever  so  long 
as  it  is  a  convex  quadrilateral.  The  logic  of  the  proof 
must  be  entirely  independent  of  the  appearance  of  the 
constructed  figure. 

The  description  of  the  figure  must  contain  all  the  con- 
ditions given  by  the  hypothesis. 

A  good  way  to  show  that  the  description  of  the  figure  is  what  really 
enters  into  the  proof,  is  to  let  one  pupil  describe  the  figure  in  words 
and  each  of  the  others  draw  a  figure  of  his  own  to  correspond  to  that 
description.  The  proof  must  then  be  such  as  to  apply  to  every  one  of 
these  figures  though  there  may  not  be  two  of  them  exactly  alike. 

396.  EXERCISES. 

1.  Every  word  in  the  language  is  defined  in  the  dictionary.  How 
is  this  possible  in  view  of  what  has  been  said  about  the  impossibility 
of  defining  every  word  ? 

2.  Can  we  determine  experimentally  whether  or  not  the  space  in 
which  we  live  satisfies  the  parallel  line  axiom  (§  96)  ? 

3.  Can  we  determine  experimentally  whether  or  not  there  can  be 
more  than  one  straight  line  through  two  given  points? 

4.  Which  theorems  of  Chapter  I  are  found  by  direct  proof  and 
which  by  indirect  proof  ? 

5.  If  two  triangles  have  two  angles  of  the  one  equal  to  two  angles 
of  the  other,  and  also  any  pair  of  corresponding  sides  equal,  the 
triangles  are  congruent. 

6.  If  two  triangles  have  two  sides  of  the  one  equal  to  two  sides  of 
the  other,  and  also  any  pair  of  corresponding  angles  equal,  the  triangles 
are  congruent  in  all  cases  except  one.  Discuss  the  various  cases 
according  as  the  given  equal  angles  are  greater  than,  equal  to,  or  less 
than  a  right  angle,  and  are,  or  are  not,  included  between  the  equal 
sides,  and  thus  discover  the  exceptional  case. 

7.  State  a  theorem  on  the  congruence  of  right  triangles  which  is 
included  in  the  preceding  theorem. 


224 


PLANE  GEOMETBY, 


8.  What  theorems  of  Chapter  I  on  parallel  lines  can  be  proved 
without  the  parallel  line  axiom  ? 

9.  What  theorems  of  Chapter  I  on  parallelograms  can  be  proved 
without  the  parallel  line  axiom  ? 

10.  What  regular  figures  of  the  same  kind  can  be  used  to  exactly 
cover  the  plane  about  a  point  used  as  a  vertex  ? 

11.  What  combinations  of  the  same  or  different  regular  figures  can 
be  used  to  exactly  cover  the  plane  about  a  point  used  as  a  vertex  ? 

12.  Suppose  it  has  been  proved  that  the  base  angles  of  an  isosceles 
triangle  are  equal  but  that  the  converse  has  not  been  proved. 

On  this  basis  can  it  be  decided  whether  or  not  the  base  angles  are 
equal  by  simply  measuring  the  sides  ? 

Can  it  be  decided  on  the  same  basis  whether  or  not  the  sides  are 
equal  by  simply  measuring  the  base  angles?    Discuss  fully. 

13.  The  sum  of  the  three  medians  of  a  triangle  is  less  than  the  sum 
of  the  sides.     See  Ex.  34,  p.  83. 

14.  The  sum  of  the  three  altitudes  of  a  triangle  is  less  than  the 
sum  of  the  sides. 

15.  A  triangle  is  isosceles  (1)  if  an  altitude  and  sm  angle-bisector 
coincide,  (2)  if  an  altitude  and  a  median  coincide,  (3)  if  a  median 
and  an  angle-bisector  coincide. 

16.  If  two  sides  of  a  triangle  are  unequal,  the  medians  upon  these 
sides  are  unequal  and  also  the  altitudes. 

17.  An  isosceles  triangle  has  two  equal  altitudes, 
two  equal  medians,  and  two  equal  angle-bisectors. 

18.  ABCD  is  a  square  and  the  points  E,  F,  G,  H, 
are  so  taken  that  AE  =  AH  =  CF  =  CG.  Prove 
that  EFGH  is  a  rectangle  of  constant  perimeter, 
whatever  the  length  of  AE. 

19.  The  bisectors  of  the  exterior  angles  of 
a  parallelogram  form  a  rectangle  the  sum  of 
the  diagonals  of  which  is  the  same  as  the 
sum  of  the  sides  of  the  parallelogram. 

20.  Prove  that  the  perpendicular  bisectors 
of  the  sides  of  a  polygon  inscribed  in  a  circle 
meet  in  a  point.  Use  this  theorem  to  show 
that  the  statement  is  true  of  any  triangle. 


REVIEW  AND   FURTHER   APPLICATIONS.  225 

21.  The  bisectors  of  the  exterior  angles  of  any  quadrilateral  form 
a  quadrilateral  whose  opposite  angles  are  supplementary. 

Definition.  In  a  polygon  of  n  sides  there 
are  n  angles  and  hence  2  n  parts.  Parts  a, 
/.  A^h^  Z.  B,  etc.,  are  said  to  be  consecutive  if 
a  lies  on  a  side  of  Z  ^,  5  lies  on  the  other 
side  of  Z^,  and  also  on  a  side  of  ZjB,  etc. 

22.  Is  the  following  proposition  true  ?  If  in  two  polygons  each 
of  n  sides  2n  —  3  consecutive  parts  of  one  are  equal  respectively  to 
2n  —  3  consecutive  parts  of  the  other,  the  polygons  are  congruent. 

Suggestion.  Try  to  prove  this  proposition  for  n  =  3,  fhen  for 
n  =  4,  and  finally  for  the  general  polygon. 

23.  What  theorems  on  the  congruence  of  triangles  are  included  in 
the  preceding  proposition  ? 

A  proposition  may  be  proved  not  true  by  giving  one  ex- 
ample in  which  it  does  not  hold. 

24.  Is  the  following  proposition  true?  If  in  two  polygons  each  of 
n  sides  2  n  —  3  parts  of  one  are  equal  respectively  to  2  n  —  3  correspond- 
ing parts  of  the  other,  the  polygons  are  congruent  provided  at  least 
one  of  the  equal  parts  is  a  side. 

25.  Given  three  parallel  lines,  to  construct  an  equilateral  triangle 
whose  vertices  shall  lie  on  these  lines. 

Solution.  liCt  P  be  any  point  on  the 
middle  line  l^.  Draw  PC  and  PA,  making 
an  angle  of  60°  with  l^.  Through  the  points 
A,  P,  C  construct  a  circle  meeting  /g  in  -6. 
Then  ABC  is  the  required  triangle. 

Suggestion  for   Proof.     Comparer!  A~' 

BPA  and  BCA  also  A  BPC  and  BA  C. 

26.  Show  how  to  modify  the  construction  of  the  preceding  example 
so  as  to  make  ABC  similar  to  any  given  triangle. 

27.  If  tangents  are  drawn  to  a  circle  at  the  extremities  of  a 
diameter  and  if  another  line  tangent  to  the  circle  at  P  meets  these 
two  tangents  in  ^  and  5  respectively,  show  that  AP  •  PB  =  r% 
where  r  is  the  radius  of  the  circle. 


G-'- 

— -. 

h 

h 

i 

y 

1 

h 

226  PLANE  GEOMETRY. 

LOCI  CONSIDERATIONS. 

397.  Two  methods  are  available  to  show  that  a  certain 
geometric  figure  is  the  locus  of  points  satisfying  a  given 
condition. 

First  method : 

Prove  (a)  Every  point  satisfying  the  condition  lies  on  the  figure. 

(b)  Every  point  on  the  figure  satisfies  the  condition. 

Second  method : 

Prove  (a')  every  point  not  on  the  figure  fails  to  satisfy  the  condition. 

(6')  ^very  point  on  the  figure  satisfies  the  condition. 

The  first  of  the  methods  is  more  direct  and  usually  more 
simple.     See  §  127. 

The  second  is  likely  to  lead  to  proofs  that  are  not 
general. 

PROBLEMS  AND  APPLICATIONS. 

1.   ^5  is  a  fixed  segment  connecting  two  parallel  lines  and  per- 
pendicular to  each  of  them.     Find  the  locus  of  the 
vertices  of  all  isosceles  triangles  whose  common 


B'      B 

_._Q^:1 

— - 

base  is  AB.     Is  the  middle  point  of  AB  a  part  of    — ^C.- -rz>— 

this  locus  ?  ""-I^^".- 

A 

2.  If  in  Ex.  1  AB  is  allowed  to  move  always 

remaining  perpendicular  to  the  given  lines,  and  if  ABC  is  any  tri- 
angle remaining  fixed  in  shape,  find  the  locus  of  the 
point  C. 

3.  Find  the  locus  of  the  centers  of  all  parallelo- 
grams which  have  the  same  base  and  equal  altitudes. 

4.  Find  the  locus  of  the  centers  of  parallelograms 
obtained  by  cutting  two  parallel  lines  by  parallel  secant  lines. 

5.  Find  the  locus  of  the  vertices  of  all  triangles  which  have  the 
same  base  and  equal  areas. 

6.  Find  the  locus  of  a  point  whose  distances  from  two  intersect- 
ing lines  are  in  a  fixed  ratio. 

Note  that  the  whole  figure  is  symmetrical  with   respect  to   the 
bisector  of  the  angle  formed  by  the  two  given  lines. 


BEVIEW  AND  FURTHER  APPLICATIONS. 


227 


7.   Find  the   locus  of   a  point  such,  that  the  difference  of  the 
squares  of  its  distances  from  two  fixed  points  is  a  constant. 

Note  that  the  locus  must  be  symmetrical  with 
respect  to  the  perpendicular  bisector  of  the  segment 
connecting  the  two  given  points  A  and  B. 


Show  that  in  the  figure  AP'^  -  FB^  =  AC^-  CB^- 


A'^r^rzzi 


-7*B 


1 

'i 

p 

B 

^■o 

D 

8.  The  lines  l^  and  l^  meet  at  right  angles  in  a 
point  A.     0  is  any  fixed  point  on  l^.     Through  0  draw  a  line  meet- 
ing Zj  in  B.     P  is  a  varying  point  on  this 
line  such  that  OB  -  OP  is  fixed.     Find  the 
locus  of  P  as  the  line  swings  about  0  as  a 
pivot. 

Suggestion.  Draw  PD  ±  to  BP. 
Show  that  BO  '  OP  =A0  •  OD.  Hence 
we  obtain  a  set  of  right  triangles  whose 
common  hypotenuse  is  OD.     Find  the  locus  of  the  vertex  P. 

9.  Find  the  locus  of  all  points  the  sum  of  whose  distances  from 
two  intersecting  lines  is  equal  to  a  fixed  segment  a. 

Suggestion.  If  AB  and  CB  are  the  given  lines, 
investigate  points  on  the  base  AC  of  the  isosceles 
triangle  ABC  in  case  the  perpendicular  from  C  on 
AB  is  equal  to  a. 

10.  The  figure  ABCDEF  is  a  regular  hexagon. 
AG,  BH,  etc.,  are  equal  segments  bisecting  the 
angles  A,  B,  etc. 

(a)  Prove  GHWAB. 

(b)  Prove  that  the-  inner  figure  is  a  regular 
hexagon. 

Suggestion.  Show  ABHG  congruent  to  the 
other  similar  parts  by  superposition. 

(c)  Find  how  many  degrees  in  ZA  GH  and  Z  BHG,  as  needed  by 
a  carpenter  in  making  a  hexagonal  frame. 

11.  The  figure  ABCD  is  a  parallelogram.     The  points  E,  F,  G,  H 
are  taken  so  that  AE  =  CG  and  AH  =  CF. 

Vrowe  A  FCG  ^  A  AEH  and  A  EOH^  A  FOG.  D      G  _C 

This  figure  is  found  in  an  old  Roman 
pavement  in  Sussex,  England. 


228 


PLANE  GEOMETRY. 


12.  The  figure  A  BCD  is  a  square.     AE  =  BF=  CG-  DH,  and 
Ey,  Fz,  Giv,  Hx  are  so  drawn  that  Zl=Z2  =  Zo=:Z4. 

(a)  Find  A  EyF,  FzG,  etc. 

Suggestion.     Through  B  draw  a  line  parallel  to 
Fz  and  make  use  of  the  A  thus  formed. 

(h)  Prove  EBFy  ^  FCGz  =  GDHw  =  HAEx. 
(c)  What  kind  of  figure  is  :r?/zt<;?    Prove. 

See  the  accompanying  design. 

13.  In  a  square  ABCD  diagonals  and  diameters 
are  drawn  as  shown  in  the  figure.  (A  line  connect- 
ing the  middle  points  of  opposite  sides  of  a  quadri- 
lateral is  a  diameter.)  The  points  K,  Z,  N  are  laid 
off  so  that  MN=  MK  =  ML.  The  small  triangles 
on  the  other  sides  are  constructed  congruent  to  KLN.  Through  the 
vertices  of  the  triangles  lines  EF,  EG,  etc.,  are  drawn  parallel  to  the 
sides  of  the  given  square. 


\ 

/ 

^ 

s 

/ 

N 

Parquet  Pattern. 


D 

n 

\H\ 

/ 

/ 

\ 

\ 

/ 

/ 

P 

/ 
K 

/ 

\ 

F 

/ 

/ 

\ 

\ 

A.      K  M  L      B 


Tile  Pattern. 


Prove  that  (a)  AKNE,  LBFN,  etc.,  are  congruent  parallelograms 
and  hence  that  EF,  EG,  etc.,  meet  in  points  on  the  diagonals.  (See 
Ex.  6,  §  123.) 

(b)  ABFE^BCGF. 

(c)  EFGH  is  a  square. 

(d)  KN  and  QP  lie  on  the  same  straight 
line. 

14.  The  figure  ^5 CD  is  a  square.  PQ 
and  RS  are  diameters.  The  points  E,  F,  G, 
H,  :.  are  so  taken  that  AE  =  BE  =  BG 
=  HC  =  ....      Also  ON^OK=OL  =  OM. 


^ 

w^ 

^ 

^ 

kS 

^ 

^ 

^ 

^ 

^ 

8^ 

^ 

OF 

REVIEW  AND  FURTHER  APPLICATIONS.  229 

(a)  Prove  that  A  ENF  ^:  A  GKH. 

(b)  Prove  that  TMONEA  =  FNOKGB. 

(c)  How  many  axes  of  symmetry  has  the  figure 
TAENFBGKH  .-."i 

(d)  Has  this  figure  a  center  of  symmetry  ? 

See  the  four  squares  in  the  accompanying 
design. 

15.  A  cross-over  track  is  constructed  as  shown  in  the  figure. 
The  rails  of  the  curved  track  are 
tangent  to  those  of  the  main  line  at 
A,  B,  F  QXidi  G.  The  curves  are  tan- 
gent to  each  other  at  D  and  E.  The 
arcs  GE  and  BD  have  equal  radii  as 
have  the  arcs  A  E  and  DF.  C^  is  the 
center  of  the  arcs  GE  and  FD  and  C2 
is  the  center  of  the  arcs  BD  and  A  E. 

(a)  Prove  that  C^HC^  is  a  right 
triangle. 

(h)  If  the   distance   between  the 
tracks  is  h  feet  and  the  distance  be- 
tween the  rails  in  each   track  (the        c^ 
gauge)  is  a  feet,  show  that 

/2 


^^ 


y 

(2  x  +  3  a  +  2  &)2  =  ?/2  +  (2  a;  +  2  a  +  hy. 

Since  a  and  h  are  given,  this  equation  may  be  solved  for  x  in  terms 
of  y  or  for  y  in  terms  of  x. 

Hence  if  the  distance  AK  \^  known,  we  may  use  this  equation  to 
compute  the  radii  of  the  arcs  used  in  constructing  the  figure. 

On  the  other  hand,  if  the  radii  of  the  arcs'  are  known,  we  may 
compute  the  distance  A  K. 

This  is  a  very  common  problem  in  railway  construction. 
The  construction  is  also  used  in  laying  out  a  curved  street 
to  connect  two  parallel  streets. 

16.  If  in  the  preceding  problem  a  =  4  feet  8|  inches,  ft  =  9  feet, 
and  AK  =  200  feet  (an  actual  case),  find  the  radii  of  the  arcs. 

Using  the  same  values  for  a  and  b  as  in  the  preceding  example, 
find  AK  if  C^E  =  300  feet. 


230  PLANE  GEOMETBY. 

FURTHER  DATA  CONCERNING  CIRCLES. 

398.  In  Chapter  II  numerous  theorems  were  proved 
concerning  the  equality  of  arcs,  angles,  chords,  etc. 

The  three  following  theorems  involve  inequalities  of 
these  elements.  The  student  should  construct  figures  in 
each  case  and  give  the  proof  in  full. 

399.  Theorem.  In  the  same  circle  or  in  equal  circles, 
of  two  unequal  angles  at  the  center  the  greater  is  suh- 
te7ided  hy  the  greater  arc  ;  and  of  tivo  unequal  arcs  the 
greater  subtends  the  greater  angle  at  the  center. 

The  proof  is  made  by  superposition. 

400.  Theorem.  In  the  same  circle  or  in  equal  circleSy 
of  two  unequal  minor  arcs  the  greater  is  subtended  by 
the  greater  chord ;  and  of  two  unequal  chords  •  the 
greater  subtends  the  greater  minor  arc. 

The  proof  depends  upon  the  theorem  of  §  117. 

401.  Theorem,  hi  the  same  circle  or  in  equal  circles, 
of  tioo  unequal  angles  at  the  center,  both  less  than 
straight  angles,  the  greater  is  subtended  by  the  greater 
chord;  and  of  two  unequal  chords  the  greater  subtends 
the  greater  angle  at  the  center. 

402.  Problem.  On  a 
given  line-segment  as  a 
chord  to  construct  an  arc 
of  a  circle  in  ivhich  a 
given  angle  may  be  in- 
scribed. .  „ 

Construction.  Let  ^  be  the  given  angle  and  AB  the  given 
segment. 


REVIEW  AND  FURTHER  APPLICATIONS.  231 

It  is  required  to  construct  a  circle  in  which  AB  shall  be 
a  chord  and  such  that  an  angle  inscribed  in  the  arc  AMB 
shall  be  equal  to  the  given  angle  K. 

At  A  construct  an  angle  BAC  equal  to  Z  K. 

If  now  a  circle  were  passed  through  A  and  B  so  as  to 
be  tangent  to  AC  at  A,  then  one  half  the  arc  AB  would 
be  the  measure  of  the  angle  BAG.     See  §  219. 

Then  any  angle  inscribed  in  the  arc  AMB  as  Z  APB 
would  be  equal  to  ZbaC—Zk. 

Hence  the  problem  is  to  find  the  center  of  a  circle  tan- 
gent to  AC  at  A  and  passing  through  A  and  B. 

Let  the  student  complete  the  construction  and  proof. 

403.  EXERCISES. 

1.  Prove  that  the  problem  of  §  402  may  be  solved  as  follows: 
With  vl,  any  pomt  on  one  side  of  the  angle  K,  as  center  and  AB  as 
radius  strike  an  arc  meeting  the  other  side  of  the  angle  at  B.  Cir- 
cumscribe a  circle  about  the  triangle  ABK. 

2.  Show  that  from  any  point  within  or  outside  a  circle  two  equal 
line-segments  can  be  drawn  to  meet  the  circle  and  that  these  make 
equal  angles  with  the  line  joining  the  given  point  to  the  center. 

3.  Show  that  if  two  opposite  angles  of  any  quadrilateral  are  sup- 
plementary, it  can  be  inscribed  in  a  circle. 

Suggestion.      Let  ABCD  be    the   quadri-  p 

lateral  in  which  /.A  -\-  ZC  =  2  vt.A.  Z"^7^N>\ 

Pass  a  circle  through  B,  C,  and  D.     To  prove  /     /\       ^^rj 

that  A  lies  also  on  the  circle,  and  not  at  some  /      ///              \ 

inside  or  outside  point  as  A'  or  A" .  \k'/Jx             1/ 

(1)  Show  that  Z  A'  •>  Z  A  s^ndi  /.  A"  <Z  A.  ^Uj^^-^T^^ 
See  §§  217,  222.  /i>5^^^>^ 

(2)  Hence    Z.4' +  Z  C>2  rt.  J    if    A'    is  l^"^^'^ 
within  the  circle  and  Z  ^"  +  Z  C<2  rt.  A\i  A"  A" 

is  outside  the  circle,  both  of  which  are  contrary  to  the  hypothesis. 
Hence  the  fourth  vertex  must  lie  on  the  circle. 

Show  that  the  condition  that  the  polygon  is  convex  follows  from 
the  hypothesis. 


232 


PLANE  GEOMETRY, 


404.   Problem.     To  draw  a  common  tangent  to  two 
circles  ivhich  lie  wholly  outside  of  each  other. 


Construction.  Let  the  given  circles  be  0  and  0'  of  which 
the  radius  of  the  first  is  the  greater. 

Required  to  draw  a  tangent  common  to  both  circles. 

Draw  an  auxiliary  circle  with  center  o  and  radius 
equal  to  the  difference  of  the  two  given  radii  in  one  figure 
and  equal  to  the  sum  of  these  radii  in  the  other  figure. 

In  each  case  draw  a  tangent  to  this  auxiliary  circle 
from  o\  thus  fixing  the  point  D.     See  §  230. 

Draw  the  radius  OD,  thus  fixing  the  point  P. 

Draw  O^P^  II  DP,  thus  fixing  the  point  P'. 

Then  PP^  is  the  required  tangent. 

Proof :  Show  that,  in  each  case,  PP'o'd  is  a  rectangle, 
thus  making  PP'  perpendicular  to  the  radii  OP  and  o'p', 
that  is,  tangent  to  each  circle. 

Definition.  A  common  tangent  to  two  circles  is  called 
direct  if  it  does  not  cross  the  segment  connecting  the 
centers,  and  transverse  if  it  does  cross  it. 

405.  EXERCISES. 

1.  Describe  the  relative  positions  of  two  circles  if  they  have  two 
direct  common  tangents.     Also  if  they  have  only  one. 

2.  Describe  the  positions  of  two  circles  if  they  have  two  transverse 
common  tangents  j  one ;  none. 


BEVIEW  AND  FUBTHEB  APPLICATIONS, 


233 


D 

0 

^"^^ 

Tk 

G 

^ 

A 

\^^^ 

A 

B 

__4^ D 

!     yC' 

\\\/y  J 

PROBLEMS  AKD  APPLICATIONS. 

1.  Find  the  locus  of  the  middle  points  of  all  chords  of  a  circle 
drawn  through  a  fixed  point  within  it. 

2.  The  sides  AD  and  BC  of  the  square  ABCD 
are  each  divided  into  four  equal  parts  and  a  circle 
inscribed  in  the  square.  Lines  are  drawn  as  shown 
in  the  figure.  Show  that  EFGHKL  is  a  regular 
hexagon. 

This  is  tlie  construction  by  means  of 
which  a  regular  hexagonal  tile  is  cut  from  a  square  tile. 

3.  In  the  preceding  example  what  fraction  of  the  area  of  the  square 
is  covered  by  the  hexagon. 

4.  A  circle  of  constant  radius  passes 
through  a  fixed  point  A.  A  line  tangent  to 
it  at  the  point  P  remains  parallel  to  a  fixed 
line  BD.     Find  the  locus  of  P. 

Suggestion.  In  the  figure  prove  A  C'PC 
a  parallelogram.  Show  that  the  locus  consists 
of  two  circles. 

5.  The  same  as  the  preceding  except  that 
the  circle  of  constant  radius  remains  tangent 
to  a  fixed  circle  instead  of  passing  through  a 
fixed  point. 

Suggestion.  On  that  diameter  of  the  fixed 
circle  which  is  perpendicular  to  the  fixed  line 
lay  off  ^  C  =  C'P.  Prove  A  C'PC  a  parallelo- 
gram. Show  that  the  locus  consists  of  two 
circles. 

6.  In  the  square  ABCD,  AM  =  0B  =  BU  =  VC,  etc.  The  point 
E  is  the  intersection  of  the  segments  MV  and  TU,  F  is  the  inter- 
section of  OB  and  TU,  G  is  the  intersection  of  OW  

and  US,  etc.     When  these  segments  are  partially 
erased,  we  have  the  figure. 

(a)  Prove  that  THEN,  OUGF,  etc.,  are  squares. 
(6)  What  kind  of  a  figure  is  MOFE  ?     Prove, 
(c)  How  many  axes  of  symmetry  has  the  figure 
NEFGH  ■• .  ?    Has  it  a  center  of  symmetry? 


234 


PLANE  GEOMETRY. 


7.   ABCD  is  a  square  the  mid-points  of  whose  sides  are  joined. 
On  its  diagonals  points  E,  K,  P,  T  are  taken  so  that  AE  =  BK  =  CP 
=  DT.      The   construction   is   then    completed  as 
shown  in  the  figure.     {FG  and  MN  lie  in  the  same 
straight  line.) 

(a)  Prove  that  E,  K,  P,  T  are  the  vertices  of  a 
square. 

(b)  li  AB  =  Q  find  ^^  so  that  the  area  of  the 
square  EKPT  shall  be  half  that  of  the  square 
ABCD. 

(c)  If  AB  =  6,  and  if  CP  =  PS,  find  the  area 
of  the  figure  EFGHKLM  ....  Also  find  the  area 
of  the  trapezoid  BKLR. 

(d)  If  CP  =  f  CS,  find  the  area  of  the  inside 
figure  and  also  of  BKLR. 

(e)  What  fraction  of  CS  must  CP  be  in  order 
that  the  inside  figure  shall  be  half  the  square? 

(/)  If  the  inside  figure  is  |  of  the  square  and  if  AB 
find  CP. 


8  inchesj, 


8.  In  the  figure  ABCD  is  a  square.  Each 
of  its  sides  is  divided  into  three  equal  parts 
by  the  points  E,  F,  G,  H,  ■•.  The  points  E, 
X,  Y,H',  F,  X,  W,  il/;  etc.,  lie  in  straight  lines. 

(a)  If  AB  =  6  inches,  find  the  area  of  that 
part  of  the  figure  which  lies  outside  the  shaded 
band. 

(ft)  If  AB  =  6  inches,  and  if  the  width  of 
the  shaded  band  is  J  inch,  find  the  area  of  the 
band. 

(c)  JiAB  =  8  inches  and  ZP  =  J  inch,  find 
the  area  of  that  part  of  the  figure  which  lies 
inside  the  band. 

(d)  If  AB  =  8  inches,  what  must  be  the 
width  of  the  band  in  order  that  it  shall  occupy 
10  %  of  the  area  of  the  whole  design  ? 

(e)  li  AB  =  a  inches  find  the  width  of  the  band  if  it  occupies  n 
per  cent  of  the  area  of  the  whole  design. 


Parquet  Flooring. 


REVIEW  AND  FURTHER  APPLICATIONS. 


235 


,P N 


scxx 


9.  By  means  of  the  accompanying  figure 
find  the  area  of  a  regular  dodecagon  whose  sides 
are  6  inches.  Notice  that  the  dodecagon  con- 
sists of  the  regular  hexagon  in  the  center,  the  six 
equilateral  triangles,  and  the  six  squares. 

Note  how  this  figure  enters  into  the  accom- 
panying tile  design. 

10.  Find  an  expression  for  the  area  of  a  regu- 
lar dodecagon  whose  sides  are  a. 

11.  Find  the  apothem  of  a  regular  dodecagon 
by  dividing  its  area  by  half  the  perimeter.  Also 
find  it  by  finding  the  apothem  of  the  hexagon 
and  then  adding  a  to  this.     Compare  results. 

12.  Find   the  radius  of   a  circle   circumscribed   about  a  regular 
dodecagon  whose  side  is  a. 

13.  If  the  accompanying  design  for  par- 
quet border  is  10  inches  wide,  find  the 
width  of  the  strips  from  which  the  small 
squares  are  made.  Also  find  this  result  if 
the  width  of  the  border  is  a. 

14.  In  the  figure  of  the  preceding 
example  find  the  dimensions  of  the  .small 
triangles  along  the  two  edges  if  the  width 
of  the  border  is  a  inches. 

15.  Show  that  the  figure  given  below  may 
be  used  as  the  basis  for  the  design  shown 
beside  it.     (See  Ex.  7,  p.  147.) 

16.  If  q,n  outside  of  each  of  the  two 
squares  shown  in  the  figure  is  6  inches,  find 
the  width  of  the  sti-ips  of  which  they  are 
made  in  order  that  each  shall  fit  closely 
into  the  corner  of  the  other? 

Suggestion.  Find  the  altitude  on 
EM  in  A  UEM  above  if  AB  =  Q 
inches. 


This    is    a    common    Arabic 
ornament. 


236 


PLANE  GEOMETRY. 


17.  In  the  figure  ABCD  is  a  square.  On 
each  side  a  triangle,  two  of  whose  sides  are 
parallel  to  the  diagonals  of  the  square,  is  so 
constructed  that  the  points  E,  F,  and  G  lie  in 
the  same  straight  line. 

(a)  Prove  that  the  triangles  are  right  isos- 
celes triangles. 

(b)  What  part  of  the  square  lies  within 
the  triangles  ? 

(c)  If  AB  =  a,  find  AE  so  that  the  tri- 
angles shall  occupy  -  of  the  area  of  the  square. 

n 

See  the  accompanying  tile  pattern. 

18.  ABCD  is  a  square  and  the  small  fig- 
ures in  the  corners  are  squares. 

(a)  Show  that  if  the  lines  intersect  as 
shown  in  the  figure  A  E  must  be  \  of  AB. 

(b)  li  AB  —  Q  inches  find  the  areas  of  the 
squares  A'B'C'D^  and  XYZW. 

19.  In  the  tile  design  show  that  the  figure 
within  the  square  is  the  same  as  that  of  ^x. 
18.  What  part  of  the  large  square  is  occupied 
by  each  shaded  part  ? 

20.  ABCD  is  a  square.  AE  =  FB  = 
BK  =  LC  =  CP  =  QD  =DR  =  SA. 

(a)  If  AB  =  a  and  if  AE  =  b,  find  the  sum  j 
of  the  areas  of  the  four  shaded  rectangles. 

(b)  If.  AB  =  8  inches,  find  ^^  so  that  the 
sum  of  these  rectangles  shall  be  ^  of  the  whole 
square.     Interpret  the  two  solutions. 

(c)  If  AB  =  a  inches,  find  AE  so  that  the 

sum  of  the  rectangles  shall  be  -  of  the  square. 

n 

21.  (a)  Show  that  in  the  accompanying  de- 
sign for  tile  flooring  the  size  of  any  one  piece 
determines  the  size  of  every  piece  in  the  figure. 

(b)  What  fraction  of  the  figure  is  occupied  by 
each  color? 


E 


D 


/¥ 

'X/V/ 

K\ 

xf 

i 

/ 

500 

J 

\ 

uA 

BEVIEW  AND  FUBTHEB  APPLICATIONS. 


237 


22.  This  tile  floor  design  is 
based  on  the  plane  figure  and 
this  in  turn  is  based  on.  the 
figure  of  Ex.  17  where 


AE  = 


AB 


> 


< 


What  part  of  the  whole  design     A   E 
is  occupied  by  tiles  of  the  various  colors  ? 

23.  ABCD  is  a  square.  Lines  are  drawn 
parallel  to  the  sides  and  intersecting  as  shown 
in  the  figure.  , 

(a)  Show  how  the  sides  of  the  square  must 
be  divided  in  order  to  make  the  lines  intersect 
as  they  do. 

(b)  If  AB  =  Q  inches,  find  the  areas  of  the 
squares  SXZQ  and  RTYP. 

(c)  Find  these  areas  M  AB  =  a. 

(d)  Show  that  the  outline  figure  is  the  basis 
of  the  tile  design  here  given. 

24.  ABCD  and  A'B'C'D'  are  equal  squares, 
rounded  by  strips  of  equal  width. 

(a)  liAB  =  a,  find  AF,  FE,  and  EBl 

(b)  Find  the  width  of  the  strips  if  their 
outer  edge  passes  through  the  points  A,  B,  C,  D, 
when  AB  =  S  inches!     Also  when  AB  =  a. 

(c)  If  AB  =  a,  find  A'H  and  hence  KL. 

(d)  If  the  width  of  the  strip  is  1  inch,  find 
KL. 

(e)  If  the  width  of  the  strip  is  b,  find  KL. 
(/)  If  KL  =  c,  find  the  width  of  the  strip. 
When  the  width  of  the   parquet  floor  border 

shown  in  the  figure  is  given,  (/)  is  the  problem 
one  needs  to  solve  to  know  how  wide  a  dark  strip 
to  use.     Compare  Ex.  16,  page  235. 


M      L 


N 


XX 

X 

X 

X 

X 

X 

X 

K 


H 


A'B'C'D'  is  sur- 


238  PLANE  GEOMETRY. 

THE   INCOMMENSURABLE   CASES. 

406.  We  have  seen,  in  §  234,  that  there  are  segments 
which  are  incommensurable ;  that  is,  which  have  no  com- 
mon unit  of  measure,  —  for  instance,  the  side  and  the 
diagonal  of  a  square. 

For  practical  purposes  the  lengths  of  such  segments  are 
approximated  to  any  desired  degree  of  accuracy,  and  their 
ratios  are  understood  to  be  the  ratios  of  these  approximate 
numerical  measures.     See  §§  238-240. 

All  theorems  involving  the  ratios  of  incommensurable 
segments,  and  the  lengths  and  areas  of  circles,  have  thus 
far  been  proved  only  for  such  approximations,  and  these 
are  quite  sufficient  for  any  refinements  of  measurement 
which  it  is  possible  to  make. 

But  for  theoretical  purposes  it  is  important  to  consider 
these  incommensurable  cases  further,  just  as  in  algebra  we 
not  only  approximate  such  roots  as  V2,  V3,  V5,  etc.,  but 
we  also  deal  with  these  surds  as  exact  numbers. 

For  instance,  in  such  an  operation  as 

(V3  +  >/2)(  V3  -  \^)  =  3  -  2  =  1. 

While  the  length  of  the  diagonal  of  a  unit  square  cannot 
be  expressed  as  an  integer  or  as  a  rational  fraction,  that  is, 
the  quotient  of  two  integers,  we  nevertheless  think  of  such 
a  segment  as  having  a  definite  lengthy  or  what  is  the  same 
thing,  a  definite  ratio  with  the  unit  segment  forming  the 
side  of  the  square. 

E.g.   If   d  is  the  diagonal   of  the  square  whose  side   is  1,   then 

d^=l  -\-l  ov  d  =  V2.     Now  suppose  y/2  =  - ,  a  fraction  in  its  lowest 

2  ^ 

terms.      Then  2  =^,  a  fraction  also  in   its   lowest  terms.     Bat   a 

fraction  in  its  lowest  terms   cannot  be  equal  to   2.     Hence   V2   is 
neither  an  integer  nor  the  quotient  of  two  integers. 


REVIEW  AND  FURTHER   APPLICATIONS.  239 

.  407.  The  following  axiom  and  that  of  §  409  -enQ  funda- 
mental in  the  consideration  of  incommensurable  line-seg- 
ments. 

Axiom  X.  Every  line-segment  ab  has  a  definite  le?igth, 
which  is  greater  tha7i  AC  if  c  lies  between  A  and  b. 

The  length  of  a  line-segment  is  in  every  case  a  number, 
which  is  rational  (an  integer  or  the  quotient  of  two  inte- 
gers) in  case  the  segment  is  commensurable  with  the  unit 
segment,  but  which  otherwise  is  irrational.  Under  the 
operations  of  arithmetic  these  irrational  numbers  obey  the 
same  laws  as  the  rational  numbers. 

The  length  of  a  line-segment  is  often  called  its  numeri- 
cal measure. 

408.  The  exact  ratio,  or  simply  the  ratio,  of  two  line- 
segments  is  the  quotient  of  their  numerical  measures, 
whether  these  are  rational  or  irrational.  That  is,  every 
such  ratio  is  a  number. 

It  is  obvious  that  a  segment  may  be  constructed  whose 
length  is  any  given  rational  num|)er.  We  have  also  seen 
how  to  construct  with  ruler  and  compasses  segments  whose 
lengths  are  certain  irrational  numbers,  such  as  V^,  VS, 
V5,  etc.    See  Ex.  3,  §  295. 

409.  We  now  assume  the  following 

Axiom  XI.  For  any  given  number  k  there  exists  a 
line  segment  whose  length  is  K, 

This  does  not  imply  that  it  is  possible  by  means  of  the 
ruler  and  compasses  to  construct  a  segment  whose  length  is 
any  given  irrational  number.  For  instance,  we  cannot 
thus  construct  a  segment  whose  length  is  V2. 

We  now  prove  the  fundamental  theorem  on  the  propor- 
tionality of  sides  of  triangles. 


240  PLANE  GEOMETRY. 

410.  Theoeem.  a  line  parallel  to  the  hase  of  a  tri- 
angle, and  meeting  the  other  two  sides,  divides  these  sides 
proportionally. 

Given  A  ABC  with  DE II AB^  /A 

-,  CD      CE  /         V 

To  prove        —  =  —  jj/         \q 

CA       CB  jy Ax. 

Proof:     Whether    CD   and    CA    are       /  \ 

At  — ^5 

commensurable   or   not,  we   know  by 

§§  407, 408  that  —  and  —  are  definite  numbers.    We  prove 

\yA  (yJj 

that  these  numbers  cannot  be  different. 

^J^.     .  CD  ^  CE 

t  irst,  suppose  —  < 

CA       CB 

Take  F  between  C  and  E  so  that  —  = Ax.  XI  (1) 

CA       CB 

Divide  CB  into  equal  parts,  each  less  than  EF,  Then 
at  least  one  of  the  division  points,  as  Gf,  lies  between 
E  and  F.     Draw  GH  II  AB. 

Since  CG  and  CB  are  commensurable,  we  have,  by  §  243, 

CH^CG  2) 

CA       CB  ^  ^ 

Dividing  (1)  by  (2),  we  have  —  =  -^.    But  this  can- 

CH      CG 

not  be  true,  since  CD  >  CH  and  CF  <  CG. 

Hence  —  cannot  be  less  than  — .  Why? 

CA  CB 

CD 

Secondly,  prove  in  the  same  manner  that  —  cannot  be 

CE  ^^ 

greater  than  — .     Hence,  since  the  one  is  neither  less  than 
CB 

nor  greater  than  the  other,  these  ratios  must  be  equal. 

The  following  treatment  of  incommensurable  arcs  and 

angles  is  exactly  similar  to  the  above. 


REVIEW  AND  FUBTHEB  APPLICATIONS.  241 

411.  Axiom  XII.  Any  given  arc  ab  has  a  definite 
ratio  ivith  a  unit  arc,  which  is  greater  than  that  of  an 
arc  AC,  if  c  lies  on  the  arc  ab. 

412.  Axiom  XIII.  Any  given  angle  abc  has  a  defi- 
nite ratio  ivith  a  imit  angle,  which  is  greater  than 
that  of  abb  if  bd  lies  within  the  angle  abc. 

413.  Theorem.  In  the  same  or  equal  circles  the 
ratio  of  two  central  angles  is  the  same  as  the  ratio  of 
their  intercepted  arcs. 


E 


Outline  of  Proof :    We  show  that  in  the  figure ^ 

arc  AB  ^^  "^ 

can  neither  be  less  than  nor  cfreater  than . 

arc  CD 

o „^  Z  AOB    ^  arc  AB 

suppose < . 

Z  CO'd      arc  CD 

rpi        .1      „       .-,    ,  Z.AOB      arc  AE  ^^>, 

Then  take  E  so  that —  = .  (1) 

Z  CO'd      arc  CD 

Divide  arc  CD  into  equal  parts  each  less  than  arc  EB. 
Lay  oif  this  unit  arc  successively  on  AB  reaching  a  point 
F  between  E  and  B.  Then  arcs  AF  and  CD  are  commen- 
surable and  by  a  proof  exactly  similar  to  that  of  §  243, 
making  use  of  §  199,  we  can  show  that 
Z  AOF  _  arc  AF 
Z  CO'd  ~  arc  CD ' 
Complete  the  proof  as  in  §  410. 


(2) 


242 


PLANE  GEOMETRY. 


414.  Axiom  XIV.  Any  given  rectangle  with  base  h 
and  altitude  a  has  a  definite  area  ivhich  is  greater  than 
that  of  another  rectangle  ivith  base  b'  and  altitude  a' 
if  a^a  and  b>V  or  if  a>  a  and  b  >  V. 

415.  Theorem.  Tlie  area  of  a  rectangle  is  the  prod- 
uct of  its  base  and  altitude. 


A  B 

Proof  :  Denote  the  base  and  altitude  by  b  and  a,  respec- 
tively, and  the  area  by  A. 

Suppose  A  <  ah,  and  let  a'  be  a  number  such  that  A  =  a'b. 
Lay  off  BE=:  a'. 

Consider  first  the  case  where  b  is  commensurable  with 
the  unit  segment  and  a  is  not. 

Divide  the  unit  segment  into  equal  parts  each  less  than 
CE  and  lay  off  one  of  these  parts  successively  on  BC  reach- 
ing a  point  F  between  E  and  C. 

Denote  the  length  of  BFhj  a",  and  draw  FF'  II  AB. 

Then  by  §  307  the  area  of  ABFS^  is  a"b. 

By  hypothesis  A  =  a'b,  but  a'b  <  a"b  smce  a'  <  a" . 

Hence  A  <  a"b.  (1) 

But  by  Ax.  XIV  A  >  a"b.  (2) 

Hence  the  assumption  that  A<ab  cannot  hold. 

In  the  same  manner  prove  that  the  A> ab  cannot  hold. 

The  proof  in  case  both  sides  are  incommensurable  with 
the  unit  segment  is  now  exactly  like  the  above  and  is  left 
to  the  student. 


REVIEW  AND  FURTHER  APPLICATIONS.  243 

416.  Axiom  XV.  A  circle  has  a  definite  length  and 
incloses  a  definite  area  ivhich  are  greater  than  those  of 
any  inscribed  polygon  and  less  than  those  of  any  cir- 
cumscribed polygon. 

417.  Theoeem.  For  a  giveji  circle  and  for  any 
number  k,  however  small,  it  is  possible  to  inscribe  and 
to  circumscribe  similar  polygons  such 
that  their  perimeters  or  their  areas 
shall  differ  by  less  than  k. 

Proof  :  First,  Let  p  and  p'  be  the 
perimeters  of  two  similar  polygons,  the 
first  circnmscribed  and  the  second  in- 
scribed, and  let  a  and  a'  be  their  apothems. 

Then  E  =  ±.,ovP:zPL^1^.  §245 

Hence 

Now  a—  a'  can  be  taken  as  small  as  we  like. 

Hence  p  •  — ^-~  can  be  made  as  small  as  we  please ;  that 
^         a' 

is,  p—p'  can  be  made  smaller  than  any  given  number  K. 
Second,  letting  P  and  p'  be  the  areas  of   the  circum- 
scribed   and    inscribed   polygons    respectively,    we   have 

P      a^ 

—f  =  — ^,  and  the  proof  proceeds  exactly  as  before. 

418.  Since  the  length  C  of  the  circle  is  greater  than  p' 
and.  less  than  p,  it  follows  that  C  is  thus  made  to  differ 
from  either  p  or  p'  by  less  than  K. 

And  since  the  area  A  of  the  circle  is  greater  than  p' 
and  less  than  P,  it  follows  that  A  is  made  to  differ  from 
either  P  or  p'  by  less  than  K. 


244  PLANE  GEOMETRY. 

419.  Theorem.  The  lengths  of  two  circles  are  in  the 
same  ratio  as  their  radii 

Proof:    Let  c  and  </  be  the  lengths  of  two  circles  whose 

centers  are  0  and  O'  and  whose  radii  are  r  and  r'. 

or  c 

We  shall  prove  that  —  =  —by  showing  that—  is  neither 

r 
less  than  nor  greater  than  — . 

c      v  v 

First,  suppose  that  —  <  —  or  c<c'  -—. 
c       V  r 

r 
And  let  c  differ  from  c'  -  -rhj  some  number  K. 

r 

Now  circumscribe  a  regular  polygon  P  about  O  o  with 

perimeter  p  such  that 

I><c^-^.  '   (1) 

This  is  possible  since  f  can  be  made  to  differ  from  c  by 
less  than  K  (§  418). 

Also  circumscribe  a  polygon  p'  similar  to  P  about  O  d 
with  perimeter^'. 

Then  -^  =  — -,  or  »  =  »^--7. 

p'      r         sr      J.      ^ 

r' 
But p'  > c'  and  hence  p>c'  ■—,  (2) 

c      r 
Hence  the  supposition  that  —  <—  leads  to  the  contra- 

c       r 

diction  expressed  in  (1)  and  (2)  and  is  untenable. 

o       r   • 
Now  prove  in  same  way  that  —  ^  —  is  untenable. 

420.  Theorem.  The  areas  of  two  circles  are  in  the 
same  ratio  as  the  squares  of  their  radii. 

Using  §§  348  and  418,  the  proof  is  exactly  similar  to 
that  of  the  preceding  theorem. 


REVIEW  AND  FURTHER  APPLICATIONS. 


245 


PROBLEMS  AND  APPLICATIONS. 

1.  A  billiard  ball  is  placed  at  a  point  P  on  a  billiard  table.  In 
■what  direction  must  it  be  shot  in  order  to  return  to  the  same  point 
after  hitting  all  four  sides  ? 

(The  angle  at  which  the  ball  is  reflected  from  a 
side  is  equal  to  the  angle  at  which  it  meets  the  side, 
that  is,  Zl  =  Z2,  and  Z3  =  Z4.) 

Suggestion,  (a)  Show  that  the  opposite  sides  of  the  quadrilateral 
along  which  the  ball  travels  are  parallel. 

(l>)  If  the  ball  is  started  parallel  to  a  diagonal  of  the  table,  show 
that  it  will  return  to  the  starting  point. 

2.  Show  that  in  the  preceding  problem  the  length  of  the  path 
traveled  by  the  ball  is  equal  to  the  sum  of  the  diagonals  of  the  table. 

3.  Find  the  direction  in  which  a  billiard  ball  must  be  shot  from 
a  given  point  on  the  table  so  as  to 
strike  another  ball  at  a  given  point 
after  first  striking  one  side  of  the 
table. 

Suggestion.  Construct  BE  ±  to 
that  side  of  the  table  which  the  ball 
is  to  strike  and  make  ED  =  BE. 

4.  The  same  as  the  preceding  problem  except  that  the  cue  ball  is 
to  strike  two  sides  of  the  table  before  striking  the  other  ball. 

Suggestion.     B'E'  =  E'D',  D'H  =  HF. 

5.  Solve  Ex.  4,  if  the  cue  ball  is  to  strike  three  sides  before  strik- 
ing the  other  ball,  —  also  if  it  is  to  strike  all  four  sides. 

6.  In  the  figure  ABCDEF  is  a  regular  hexagon. 
Prove  that :  (a)  AD,  BE,  and  CF  meet  in  a  point. 
(6)  ABCO  is  a  rhombus. 

(c)  The  inner  circle  with  center  at  0  and  the 
arcs  with  centers  at  A,  B,  C,  etc.,  have  equal  radii. 

(d)  The  straight  line  connecting  A  and  C  is 
tangent  to  the  inner  circle  and  to  the  arc  with 
center  at  B. 

(e)  The  centers  of  two  of  the  small  circles  lie  ^ 
on  the  line  connecting  A  and  C. 

(/)  Find  by  construction  the  centers  of  the  small  circles. 


F 


E.^ -.D 


^^B 


246 


PLANE  GEOMETRY. 


7.  In  the  figure  EG  and  FH  are  diameters  of  the  square  ABCD. 
On  the  diagonals,  points  K,  U,  V,  W 
are  laid  off  so  that  AK  =  BU  =  CV 
=  DW.  Also  LM  =  NP  =  RS  =  etc. 
EN  and  SF  are  in  the  same  straight 
line,  and  so  on  around  the  figure. 
Prove  that : 

(a)  KUVW  is  a  square. 

(b)  AKME   and  ENUB  are    equal 
trapezoids. 

(c)  L,  0,  T  lie  in  a  straight  line.  • 

(d)  The  four  heavy  six-sided  figures 
are  congruent. 

AO 


a,AK 


and  ML 


(e)  JiAB 

KL,  find  the  areas  of  the  figures  KUVW, 
AKME,  LOPNEM. 

(/)  Find  the   areas   required   under 

(e)  \tAK  =  —  and  LM  =  m-  LK. 

n 

AO 
(g)  li  AK  =  -T- ,  what  is  the  length  of  ML  if  the  four  heavy  figures 

occupy  half  the  square  ABCD'i 

8.  Find  a  side  of  a  regular  octagon  of  radius  r.     See  Ex.  3,  p.  146. 

9.  Find  the  area  of  a  regular  octagon  of  radius  r. 

10.  In  the  figure 
ABCD-'  is  a  regular 
octagon  inscribed  in 
a  circle  of  radius  r. 
Find  the  area  of  the 
triangle  ABC. 

Suggestion.  Find 
first  the  altitude  on  ^  C 
in  the  /\  AOC  and  thus 
the  altitude  of  A  ^5C. 

11.  In  the  same  fig- 
ure the  lines  CH  and  A  D  are  drawn  meeting  at  K.     Prove  that  A  BCK 
is  a  parallelogram  and  find  its  area  if  the  radius  of  the  circle  is  r. 


® 

s 

m 

F^-~-"-  G 


REVIEW  AND  FURTHER  APPLICATIONS. 


247 


12.  In  the  parquet  floor  design  given  with  Ex.  10,  the  darker  parts 
are  parallelograms  constructed  as  under  Ex.  11.  What  part  of  the 
area  is  of  white  wood  ? 

13.  In  the  figure  ABCD  and  A'B'C'D'  are  equal  squares,  placed  as 
shown.  Lines  are  drawn  through  A'^  B',  C,  and  D'  parallel  to  AB, 
BCj  CD,  and  DA,  forming  a  quadrilateral  EFGH. 


(a)  Prove  EFGH  a      ^^ 
square. 

(&)  What    part    of 
the     large     square     is 
inclosed    by    the    out-    jj 
side  heavy  figure? 

(c)  li  AB  =  a,  find 
the  area  of  the  inside 
heavy  figure.  E 


C 


r 

D 

/\ 

a  i 

V 

^ 

>{ 

[< 

^ 

K 

A' 


F 


14.  The  design  opposite  consists  of  white 
figures  constructed  like  the  inner  figure  preced- 
ing, together  with  the  remaining  black  figures. 
What  part  of  the  figure  is  ^hite? 

15.  Show  that  the  altitude  of  an  equilateral 
triangle  with  side  s  is  -  VS. 

16.  If  the  angles  of  a  triangle  are  30°,  60^  90°, 

and  if  the  side  opposite  the  60°  angle  is  r,  show  that 

r     -         2  r     - 
the  other  sides  are  o  V3  and  -^  V3.  See  §  159,  Ex.  14. 

17.  Three  equal  circles  of  radius  2  are  inscribed 
in^a  circle  as  shown  in  the  figure.  Find  the  radius 
of  the  large  pircle. 

Suggestion.  The  center  0  of  the  large  circle 
is  at  the  intersection  of  the  altitudes  of  the  equi- 
lateral triangle  0'0"0"'.     (Why?) 

Hence  O'OD  is  a  triangle  with  angles  30°,  60°, 
90°  as  in  Ex-  16. 

Solve  this   problem    for   any  radius   r  of  the 

2  r    /— 

small  circles.   Ans.   R  =  r  -\ v  3. 

o 


►♦♦♦♦♦♦ 
►♦♦♦♦♦♦ 

►♦♦♦♦♦♦ 
►♦♦♦#♦♦ 
►♦♦♦♦♦♦ 


Design  from  the 
Alhambra. 


s/^ 

\ 

A^r 

f  \ 

248 


PLANE  GEOMETRY, 


18.  A',  B',  C,  are  the  middle  points  of  the  sides  of  the  equilateral 
triangle  ABC.  The  sides  of  the  triangle  A'B'C  are  trisected  and 
segments  drawn  as  shown  in  the  figure. 


A     D 


E      B 


From  Church  of  Or  San  Michele. 


If  AB  =  a, 

(a)  Prove  A"B"C"  an  equilateral  triangle. 
(6)  Find  the  area  of  A"B"C"  and  of  the  dotted  hexagon, 
(c)  Find  the  area  of  the  triangle  GFC  and  of  the  trapezoid 
DEB"  A". 


A  P 


E  B 


From  Westminster  Abbey. 


19.  In  the  figure  ABC  is  an  equilateral  triangle.  On  the  equal 
bases  PM,  MD,  DE  equilateral  arches  are  constructed,  two  of  them 
tangent  to  the  sides  of  the  triangle  at  L  and  N  respectively.  Circles 
0'  and  0"  are  each  tangent  to  a  side  of  the  triangle  and  to  two  of 
the  arches.  Circle  0  is  tangent  to  circles  0'  and  0"  and  to  both 
sides  of  the  triangle. 

(a)  liAB  =  a,  find  DE. 

Suggestion.     In  the  right  triangle  DBL  one  acute  angle  is  60°. 


REVIEW  AND  FURTHER  APPLICATIONS.  249 

(h)  Find  the  ratio  r:r',r  being  the  base  DE  of  the  arch  and  r^  the 
radius  of  the  circle  0'. 


Suggestion.  GD  =  2  DL,  GO'  =2  r>,  and  O'D  =  V(r+r'y  -r^= 
y/2  rr'  4-  r'^. 

(c)  By  what  fraction  of  a  will  the  circles  0'  and  0"  fail  to  touch 
each  other? 

(d)  Find  the  radius  of  the  circle  O. 

20.  In  the  figure  CD  =  DA  =  AG  =  GB,  and  BE  =  EA  =  AF  = 
FG.  Semicircles  are  constructed  on  the  diameters  CB,  CG,  CD,  DG, 
DB,  GB.  HG  =  KD=  CE.  Arcs  GL  and  DM  have  centers  H  and 
K  respectively. 


K    C 


From  Church  of  Or  San 
Michele. 


Circles  are  constructed  tangent  to  the  various  arcs  as  shown  in  the 
figure.  Thus  O  0"  is  tangent  to  semicircles  on  the  diameters  CB,  CG, 
and  DB.  O  0'  is  tangent  to  the  semicircles  on  the  diameters  CG  and 
DB  and  to  the  arc  DM.     Let  CB  =  a. 

(a)  Find  the  areas  of  each  of  the  six  semicircles. 

(b)  Find  the  radius  r". 

Suggestion.     EA,  EP  and  ^iV  are  known. 

(c)  Find  the  radius  r'. 

Suggestion.     Enumerate  the  known  parts  in  A  EDO'  and  FDO'. 

(d)  Fi|id  r  and  r^^. 

(e)  Having  determined  the  radii  of  the  various  circles,  show  how 
to  construct  the  whole  figure. 

(/)  What  fraction  of  the  area  of  the  whole  figure  is  occupied  by 
the  six  circles  ? 

21.  Prove  that  two  segments  drawn  from  vertices  of  a  triangle  to 
points  on  the  opposite  sides  cannot  bisect  each  other. 


250  PLANE  GEOMETRY. 

FURTHER  APPLICATIONS  OF  PROPORTION. 

421.  Definition.     A  segment  is  said   to  be  divided  in 
extreme  and  mean  ratio  by  a  point  on  it,  provided  the  ratio 
of  the  whole  segment  to  the  larger  part    ^  ^        ^ 
equals  the  ratio  of  the  larger  part  to 

the  smaller. 

E.g.  the  segment  AB  is  divided  in  extreme  and  mean  ratio  by  the 

point  C  if  ^  =  ^. 
^  AC      CB 

422.  Problem.     To  divide  a  giveri  line-segment    in 
extreme  and  mean  ratio. 

Solution.    At   one   extremity  B 
erect  BoA-AB  and  make   bo  =  \  ^-'''' 

AB.     Draw  OA.  ^^-'  \ 

On  OA  take  OB  =  OB  and  with         .--"' 


^  as  a  center  and  AD  as  a  radius   a  g  b 

draw  the  arc  DC.     Then  C  is  the  required  point  of  division. 

Proof:  Let  a  be  the  length  of  AB.     Then  OB  =-. 

Hence      j:o-  =  i^^  +  ^2  =  «2  4.  |'==  | ^2,  or  ^O  =  |  VsT 

Then        ^C  =  ^Z)  =  yio  -  DO  =  ^  V5  -  ^  =f  ("^^  ~  A 
DC  =  ^B  -  ^C  =  a  -  -^(^V5  -  1^  =  1^3  -  VS)- 


and 


Substituting  these  values  in  — -  and  — ,  and  simplifying, 

AC         CB  • 

wo  have  — and  ^"^  ~  _ 

V5  -  1  3  -  V5* 

By  rationalizing  denominators  these  fractions  are  shown 

to  be  equal,  and  hence     —  =  — • 
AC       CB 


REVIEW  AND  FURTHER  APPLICATIONS. 


251 


423.    Problem.      To  consti^uct  with  ruler  and  com- 
passes angles  0/36°  and  72°. 

Solution.    On  a  given  segment  AB^  determine  a  point  0 

such  that  i^  =  £^.     (See  §  422.) 
AC       BG 

Using  BD  —  AG  as  a  base  and  AB  as  one  leg,  construct 
the  isosceles  triangle  ABD.  ^ 

Then  Z 1  =  72°  and  Z  4  =  36°. 

Proof :  Draw  BG. 
Then  A  ABD  ~  A  DBG,  (§  259) 


since 


AB       BD 


and  Z  1  is  common . 
BC  A 


BD       BC  AC 

Hence  A  ABD  and  BDG  are  both  isosceles  (Why  ?), 
and  Z  1  =  Z  2,  Z  3  =  Z  4. 

Also  Z2  =  Z3  +  Z4  =  2Z4  =  Z1.    (Why?) 

Thus  in  A  ABD  each  base  angle  is  double  Z4, 
making  Z4  =  |  of  2  rt.  A  =  36°,  (Why?) 

and  Z1  =  2Z4  =  72°. 


424. 


EXERCISES. 


1.  Inscribe  a  regular  decagon  in  a  circle. 

Suggestion.  Construct  the  central  angle 
A  OB  =  36",  thus  determining  the  side  AB  oi  the 
decagon. 

2.  Inscribe  a  regular  pentagon  in  a  circle. 

Suggestion.     Join  alternate  vertices  of  a  decagon,  or  construct  a 
central  angle  equal  to  72°. 

3.  Inscribe  a  regular  polygon  of  fifteen  sides  in  a  circle. 
Suggestion.     Since  ^  —  J^  =  -^■^,  it  follows  that  the  arc  subtended 

by  one  side  of  a  hexagon  minus  that  subtended  by  one  side  of  a  deca- 
gon is  the  one  which  subtends  one  side  of  a  polygon  of  fifteen  sides. 


252  PLANE  GEOMETRY. 

4.  The  radius  of  an  inscribed  regular  polygon  is  a  mean  propor- 
tional between  its  apothem  and  the  radius  of  a  similar  circumscribed 
polygon. 

5.  A  circular  pond  is  surrounded  by  a  gravel  walk,  such  that  the 
area  of  the  walk  is  equal  to  the  area  of  the  pond.  What  is  the  ratio 
of  the  radius  of  the  pond  to  the  width  of  the  walk? 

6.  lla,b,c  are  three  line-segments  such  as  -  =  -  show  that  -  equals 

he  c 

the  ratio  of  the  area  of  any  triangle  described  on  a  as  a  base  to  the 
similar  triangle  described  on  i  as  a  base. 

425.  Definitions.  A  line-segment  AB  is  said  to  be 
divided  internally  in  a  given  ratio  —  by  a  point  C  lying  on 

the  segment,  li^^Vl.     See  §  252. 
^  '      CB      n  ^ 

A  line-segment  is  said  to  be  divided  externally  in  a 

given  ratio  _  by  a  point  c'  lying                          c       B           r" 
s  ^^f  

on  AB  produced,  if =  -  • 

C'B      s 

A  line-segment  AB  is  said  to  be  divided  harmonically  if 

the  points  C  and  C^  lying  respectively  on  AB  and  on  AB 

AC      AC^ 
produced,  are  such  that  —  =  --—  • 

CB       C'B 

426.  EXERCISES. 

1.  Show  that  if  AB  is  divided  harmonically  by  C  and  C,  then 
CC  is  divided  harmonically  by  A  and  B. 

2.  Show  that  the  base  of  any  triangle  is  cut  harmonically  by  the 
bisectors  of  the  internal  and  external  vertical  angles. 

3.  Show  how  to  divide  a  line-segment  externally  in  extreme  and 

AB      C'A 

mean  ratio,  that  is,  in  the  figure  below,  so  that  = . 

C'A      C'B 

Suggestion.     In  the  figure  of  §  422  produce  BA  to  a  point  C 

such  that  C'A  =A0+  OB. 

Then  use  the  method  there     ^  A  B 

used,  to  show  that  ^—  and  -^-^  each  reduces  to  • 

C'A  C'B  2 


BEVIEW  AND  FURTHER  APPLICATIONS. 


253 


427,  Theorem.  If  an  angle  of  one  triangle  is  equal 
to  an  angle  of  the  other ^  their  areas  are  to  each  other  as 
the  products  of  the  sides  including  the  equal  angles. 


A  B    a:  b'  a 

Given  A  ABC  and  A'BfC  in  which  Z.A  =  AA\ 
A  ABC  AB ' AC 


To  prove  that 


aa'b'c'    a'b'-a'g' 

Proof  :     Place  Aa'b'c'  so  that  Z  A'  coincides  with  Z  A. 


Then 

AABC  = 

Ih- 

AB  and  A . 

A'B'C'  = 

:1A'. 

A'B' 

Hence, 

A  ABC 

aa'b'c' 

\h' 

AB 
'A'B' 

h 

AB 
A'B' 

Now  show  that     —  = 

AC 

A'C' 

t 

and  hence  that       '^^^,^   = 

aa'b'c' 

AC 

A'C' 

AB 
A'B' 

AB 

A'B' 

'AC 

'A'C' 

• 

428.    Theorem.     The  square  on  the  bisector  of  an 

angle  of   a  triangle   is  equal  to  the         ^ ^\a 

product  of  the  two  adjacent  sides 
minus  the  product  of  the  segments  of 
the  opposite  side. 

Outline  of  proof:      Produce  the  bisec- 
tor of  the  given  angle  to  meet  the  cir-  E 
cumscribed  circle. 

Since  ABDC^  AAEC    AC  •  BC  ~  CD  -  CE.  (1) 

But     CD-  CE=  CDQCD  4-  DE)  =  CD^  +  CD  •  DE  (2) 

and  CD-  DE=AD'  DB.     (Why?)     (3) 

Using  (1),  (2),  and  (3),  complete  the  proof. 


254 


PLANE  GEOMETRY. 


429.  Theorem.  The  product  of  two  sides  of  a  tri- 
angle is  eqmal  to  the  product  of  the  altitude  from  the 
vertex  in  which  these  sides  meet  and  the  diameter  of  the 
circumscribed  circle. 


Outline  of  Proof :  Using  the  figure, 
show  from  the  similar  triangles  ACT) 
and  EEC  that 

AC'BC=  CE'CB. 


430.  EXERCISES. 

1.  The  areas  of  two  parallelograms  having  an  angle  of  the  one 
equal  to  an  angle  of  the  other  are  in  the  same  ratio  as  the  product  of 
the  sides  including  the  equal  angles. 

2.  Three  semicircles  of  equal  diameter 
are  arranged  as  shown  in  the  figure. 

(a)  If  ^D  =  a,  find  the  area  bounded 
by  the  arcs  AB,  BEC,  and  CA. 

(b)  If  the  area' just  found  "is  2  square 
feet,  find  ylD. 

3.  Prove  that  the  bisectors  of  the  angles  of  any- 
quadrilateral  form  a  quadrilateral  whose  opposite 
angles  are  supplementary. 

Suggestion.  Show  that  Z3+Z4  +  Z5  +  Z6 
=  2  rt.  A,  and  hence  that  Z  1  +  Z  2  =  2  rt.  ^i. 

4.  On  each  of  two  sides  of  a  given  triangle  ABC  a.s  chords  con- 
struct arcs  in  which  an  angle  of  120°  may  be  inscribed.  If  these  arcs 
meet  in  a  point  0  inside  the  triangle,  show  that  the  three  sides  of  the 
triangle  subtend  the  same  angle  from  the  point  0. 

5.  If  Z 1  +  Z  2  +  Z  3  =  4  rt.  zi,  show  how  to  find  a  point  0  within 
a  given  triangle  ABC  so  that  Z^0j!3=Z1,  ZBOC  =  Z2,  and 
ZC0A  =  Z3. 


REVIEW  AND  FURTHER   APPLICATIONS. 


255 


6.  Given  any  two  segments  AB  and  CD  and  any  two  angles  a  and 
b,  find  a  point  O  such  that  Z  A  OB  =  Za  and  Z  COD  =  Zh.  Discuss 
the  various  possible  cases  and  the  number  of  points  0  in  each  case. 

7.  li  ZA  OB  is  a  central  angle  of  a  circle  and  if  Z  CDE  is  inscribed 
in  an  arc  of  the  same  circle  and  if  ZAOB  =  Z  CDE  =  |  rt.  A,  then 
the  chords  AB  and  CE  are  equal. 


431.  Definition.  A  set  of  lines  which 
all  pass  through  a  common  point  is  called 
a  pencil  of  lines,  and  the  point  is  called 
the  center  of  the  pencil. 


432. 


EXERCISES. 


C^ 

/a:' 

A 

\  r 
/TV 

/r\ 

/A' 

IB' 

'c 

1.  The  lines  of  a  pencil  intercept  proportional 
segments  on  parallel  transversals. 

Given  three  lines  meeting  in   O  cut  by  three 
parallel  transversals. 

To  prove  that  the  corresponding  segments  are 
proportional,  that  is,  to  show  that 
AB  ^  A[R  ^  A"B" 
BC      B'C      B"C"' 

2.  Ii  two  polygons  are  symmetrical  with  respect  to  a  point,  they 
are  congruent.     (See  §  170,  Ex.  2.) 

3.  Any  two  figures  symmetrical  with 
respect  to  a  point  are  congruent. 

Suggestion.  About  the  point  0  as 
a  pivot  swing  one  of  the  figures  through 
a  straight  angle.  Then  any  point  P'  of 
the  right-side  figure  will  fall  on  its  sym- 
metrical point  P  of  the  left-side  figure. 

4.  By  means  of  the  theorem  in  the  preceding  example  show  how 
to  make  an  accurate  copy  of  a  map. 

Suggestion.  Fasten  the  map  to  be  copied  on  a  drawing  board. 
A  long  graduated  ruler  is  made  to  swing  freely  about  a  fixed  point  0, 
and  by  means  of  it  construct  a  figure  symmetrical  to  the  map  with 
respect  to  the  point  0.     How  are  the  distances  from  O  measured  off  ? 


256  PLANE  GEOMETRY. 

433.  Theorem.     If  correspo7iding   vertices   of  kvo 
polygons  lie  on  the  same  lines  of  a  pencil  and  if  they 
cut    off  proportiotial    dis- 
tances on  these  lines  from 
the   center,   then   the  poly- 
gons are  similar. 

Prove  the  theorem  first  for 
triangles. 

OB        OC        OA         -,     OA         OB  OC 

GiVen.       — 7  =  — 7  =  — 7  and  — -  =  — -  =  — -. 
ob'     oc'     oa'         oa"     ob"     oc" 

To  prove  that  A  ABC,  A'b'&,  a"b^'c"  are  similar. 
Prove   the   theorem   for   polygons   of   any   number   of 
sides. 

434.  Definition.     Any  two  figures  are  said   to   have  a 

center  of  similitude  o,  if  for  any  two  points  Pj  and  P^  the . 

lines  P^O  and  P^O  meet  the  other  figure  in  points  P\  and 

p'„  such  that 

P^O^P^^ 


P\0       P\0 

Then  P\  and  P'^  are  said  to  correspond  to  the  points 
Pj,  and  Pg. 

Thus  in  the  figure  of  §  433  O  is  called  the  center  of 
similitude  of  the  two  polygons. 

Any  two  figures  which  have  a  center  of  similitude  are 
similar.  * 

This  affords  a  ready  means  of  constructing  a  figure 
similar  to  a  given  figure  and  having  some  other  required 
property  that  is  sufficient  to  determine  it. 

Definition.  The  ratio  of  any  two  corresponding  sides  of 
similar  polygons  is  called  their  ratio  of  similitude. 


REVIEW  AND  FURTHER  APPLICATIONS.  257 

435.  EXERCISES. 

1.  Construct  a  polygon  similar  to  a  given  polygon  such  that  they 
shall  have  a  given  ratio  of  similitude. 

Suggestion.     Let  ABODE F  be  the  given   polygon   and   —  the 

n 
given  ratio  of  similitude.     Select  any  convenient  point  O,  such  that 
OA  =m.     Draw  lines   OA,  OB,  etc.     On   OA  lay  off  OA',  n  units. 
Lay  off  points  B',  C,  etc.,  so  that 

OA  ^  OB  ^  PC  ^  OP  ^  OE  ^  OF 
OA'  ~  OB'      OC      OB'      OE'      OF'' 

Prove  that  A'B'C'D'E'F'  is  the  required  polygon. 

2.  Show  how  the  preceding  may  be  used  to  ehlarge  or  reduce  a 
map  to  any  required  size. 

Suggestion.     Arrange  apparatus  as  under  Ex.  4,  §  432. 

3.  In  the  figure  A  BCD  is  a  parallelogram  whose  sides  are  of  con- 
stant length.  The  point  A  is  fixed,  while 
the  remainder  of  the  figure  is  free  to  move. 
Show  that  the  points  P  and  P'  trace  out 
similar  figures  and  that  their  ratio  of  simili- 
tude is  — ^^— . 

AD+  CP 

This  shows  the  essential  parts  of  an  in- 
strument called  the  pantograph,  which  is  much  used  by  engravers  to 
transfer  figures  and  to  increase  or  decrease  their  size.  The  point  P 
is  made  to  trace  out  the  figure  which  is  to  be  copied.  Hence  P'  traces 
a  figure  similar  to  it.  The  scale  or  ratio  of  similitude  is  regulated  by 
adjusting  the  length  of  CP. 

4.  Construct  a  triangle  having  given  two  angles  and  the  median  a 
from  one  specified  angle. 

Solution.     Construct    any  triangle   ABC  having   the  required 
angles  and   construct   a  median  AD.     Prolong   DA  to  D',  making 
AD'  =a.     Extend   BA  and  CM   and  through  D'    draw   B'C'WBC, 
making  AAB'C.     Prove  that  this  is  the  required 
triangle.  (Notice  that  yl  is  the  center  of  similitude.)  y\ 

5.  Liscribe  a  square  in  a  given  triangle  using 
the  figure  given  here.  Compare  this  method  with 
that  given  on  page  147. 


258 


PLANE  GEOMETRY. 


6.  Construct  a  circle  through  a  given  point  tangent  to  two  given 
straight  lines. 

Solution.  Let  a  and  h  be  the  given  lines  and  P  the  given  point. 
Construct  any  circle  0  tangent  to  a  and 
t.  Draw  AP  meeting  the  circle  0  in 
C  and  D.  Draw  CO  and  OD  and 
through  P  draw  lines  parallel  to  these 
meeting  the  bisector  of  the  angle 
formed  by  a  and  h  in  0'  and  0''.  Prove 
that  0  and  0"  are  centers  of  the  re- 
quired circles.  Observe  that  ^  is  a 
center  of  similitude. 

This  method  is  used  to  construct  a  railway  curve  through  a  fixed 
point  connecting  two  straight  stretches  of  road. 

7.  On  a  line  find  a  point  which  is  equidistant  from  a  given  point 
and  a  given  line. 

The  following  is  another  instance  of  the  use  of  this  very  important 
device  in  constructing  figures  that  resist  other  methods  of  attack. 
It  consists  essentially  in  first  constructing  a  figure  similar  to  the  one 
required  and  then  constructing  one  similar  to  this  and  of  the  proper  size. 


From  Westminster  Abbey. 

8.  Given  a  circle  with  radius  OE.  Construct  within  it  the  design 
shown  in  the  figure.  That  is,  the  inner  semicircles  have  as  diameters 
the  sides  of  a  regular  sixteen-sided  polygon.  Each  of  the  small  circles 
is  tangent  to  two  semicircles.  The  outer  arcs  have  their  centers  on  the 
given  circle  and  each  is  tangent  to  two  small  circles.  All  these  arcs 
and  circles  have  equal  radii. 


REVIEW  AND  FURTHER  APPLICATIONS. 


259 


Solution.  Construct  an  angle  equal  to  one  sixteenth  of  a  peri- 
gon.  Through  any  point  D'  in  the  bisector  of  this  angle  draw  a  seg- 
ment EF  perpendicular  to  the  bisector 
and  terminated  by  the  sides  of  the  angle. 
On'  EF  as  a  diameter  construct  a  semi- 
circle.    Make  B'D'  =  EF  =  B'A'  =  D'C 

=  C'A'  and  construct  the  small  figure.  E    C 

Now  draw  radii  of  the  given  circle  dividing  it  into  sixteen  equal 
parts  and  bisect  one  of  the  central  angles  by  a  radius  OA.  Construct 
Z.  DAB  =  Z  D'A'B'.  Then  ^  J5  is  twice  the  radius  of  the  required  arcs 
and  circles.     The  whole  figure  may  now  be  constructed. 

9.  Given  any  three  non-collinear  points  A',  B',  C,  to  construct  an 
equilateral  triangle  such  that  A',  B',  C  shall  lie  on  the  sides  of  the 
triangle,  one  point  on  each  side. 

Suggestion.  Through  one  of  the  points  as  A'  draw  a  line  such 
that  B'  and  C  lie  on  the  same  side  of  it. 

10.  Given  an  equilateral  triangle  A  BC,  to  construct  a  triangle  sim- 
ilar to  a  given  triangle  A'B'C  with  its  vertices  on  the  sides  of  ^i^C 

Suggestion.  Construct  an  equilateral  triangle  such  that  A',  B',  C 
lie  on  its  sides.  Then  construct  a  figure  similar  to  this  and  of  the 
required  size. 

11.  If  jo  and/>'  are  similar  polygons  inscribed  in  and  circumscribed 
about  the  same  circle,  and  if  2.s  is  a  side  of  the  circumscribed  polygon 
p',  show  that  the  difference  of  the  areas  of  p  and  p'  is  equal  to  the 
area  of  a  polygon  similar  to  these  and  having  a  radius  s. 

Suggestion.  Let  the  areas  of  the  polygons  whose  radii  are  r,  r', 
and  she  A,  A',  A".     Prove  that 

A'  -A  ^  r"^  -  r^  ^^^^  _A^  _  ^  _.  r"^  -  r^ 
A'      ~      r'2      ^"      A'  ~  r'2  ~      /^ 
Complete  the  proof. 

12.  Two  circles  are  tangent  at  A.  A 
secant  through  A  meets  the  circles  at  B  and 
C  respectively.  Prove  that  the  tangents 
at  B  and  C   are   parallel   to   each    other. 

13.  Prove  that  the  segments  joining  one  vertex  of  a  regular  poly- 
gon of  n  sides  to  the  remaining  vertices  divide  the  angle  at  that  vertex 
into  n-2  equal  parts. 


260 


PLANE  GEOMETBY. 


FURTHER   PROPERTIES  OF   TRIANGLES. 

436.  Definition.  A  segment  AB  is  said  to  be  projected 
upon  a  line  I  if  perpendiculars  from  A  and  B  are  drawn 
to  I.  If  these  meet  I  in  points  C  and  D,  then  CB  is  the  pro- 
jection of  AB  upon  I. 

437.  Theokem.  The  square  of  a  side  opposite  a7i 
acute  angle  of  a  triangle  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides  minus  twice  the  product  of 
one  of  these  sides  and  the  projection  of  the  other  ujoon  it. 


c  m 

Outline  of  Proof :     In  either  figure  let  Z  i?  be  the  given 

acute  angle,  and  in  each  case  BD  is  the  projection  of  BC 

upon  AB.     Call  this  projection  m. 

We  are  to  prove  that  5^  =  a^  -{-  e^  —  2  cr/i. 
In  the  left  figure,  b^=.h^-{.(c-  my.  (1) 

In  the  right  figure,  h^  =  h^ -\-  (^m  —  c^.  (2) 

In  either  case  W'  =  c^—  m^.  (3) 

Substitute  (3)  in  (1)  or  in  (2),  and  complete  the  proof. 
Modify  each  figure  so  as  to  draw  the  projection  of  AB  upon  BC 

and  call  this  n.     Then  give  the  proof  to  show  that  h^  =  a^  +  c"^  —  2  an. 

438.  EXERCISE. 

1.    The  area  of   a  polygon  may  be  found  by  drawing  its  longest 
diagonal  and  letting  fall  perpendiculars  upon  this 
diagonal  from  each  of  the  remaining  vertices. 

Draw  a  figure  like  the  one  in  the  margin,  only 
on  a  much  larger  scale,  measure  the  necessary 
lines,  compute  the  areas  of  the  various  parts 
(see  §  314),  and  thus  find  its  total  area. 


REVIEW  AND  FU ETHER  APPLICATIONS. 


261 


439.  Theorem.  The  square  of  the  side  opposite  an 
obtuse  angle  of  a  triangle  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides  plus  twice  the  product  of 
one  of  these  sides  and  the  projection  of  the  other  upon  it. 


A  ^  B  D 

Outline  of  Proof :  Let  Z  b  he  the  given  obtuse  angle  and 
BD  the  projection  of  BC  upon  AB.  Call  this  projection  m. 
As  in  the  preceding  theorem  show  that 

Also  modify  the  figure  so  as  to  show  the  projection  of 
AB  on  BC  and  call  this  n.     Then  show  that 

J2  =  ^2  +  (?2  _p  2  an. 

440.   Theorem.   The  sum.  of  the  squares  of  two  sides  of 
any  triangle  is  equal  to  tivice  the  square         ^ 
of   half  the   third  side  plus    twice  the     /^\^"^^^^ 
square  of  the  median  drawn  to  that  side,   a        d       b 
Suggestion.     Make  use  of  the  two  preceding  theorems. 

441.  EXERCISES. 

1.  Compute  the  medians  of  a  triangle  in  terms  of  the  sides. 

2.  Show  that  the  difference  of  the  squares  of  two  sides  of  a  tri- 
angle is  equal  to  twice  the  product  of  the  third  side  and  the  projection 
of  the  median  upon  that  side. 

3.  The  base  of  a  triangle  is  40  feet  and  the  altitude  is  30  feet. 
Find  the  area  of  the  triangle  cut  off  by  a  line  parallel  to  the  base  and 
10  feet  from  the  vertex. 


262 


PLANE  GEOMETRY. 


442.   Problem.     To  express  the  area  of  a  triangle  in 
terms  of  its  three  sides. 
c 


or 


Solution.  The  area  of  A  ABC  =  |  AB  x  CD  =  i  he. 
It  is  first  necessary  to  express  h  in  terms  of  a,  b,  c. 
We  have  b^=  0^+ e^- 2  cm,  (Why?) 


Also 


h?  =  a^  —  m^. 


(Why?) 


Hence     h^=  a^ 


a2^c'i_p\2  _  4  ^2^2  _  (^2  +  g2  _  ^2  )2 

^  (2  gg  +  a2  4-  g2  _  ^2)(2  ^6'  -  ^2  -  g2  _}_  52) 
4^2 

^  [(^  4-  e)2  -  p-\  [-62  -  (g  -  6')2] 
4  6'2 

^(a  +  e+  h){a  -\-  c  -  h){h  +  a  -  cfh  -  a  +  c) 

Now  call  a  +  b-{-  c=2s,       or  a  +  c— b  =  2s— 2  b. 
Then  a  +  c-b=2(s-b) 

b  -\-  a  —  c  =  2(s  -  e) 
b  -[-  c—  a=2{s—  a). 

Hence  K^  =  2  ^ -^C^  -  ^)  ■  2(.  -  ^)- 2(a->^)^ 

4  <?2 


or 


s  —  a)(s  —  5)(s  —  <?). 


REVIEW  AND   FURTHER  APPJ^ICATIONS.  263 

1  19      

Then       ^he  =  -c-  ^ Vs(s  - a)(s  -  5)(s  -  c). 

Hence  area  of  A  ABC  =  Vs(s  —  ^X^  —  ^X^' "~  ^)' 
443.    Peoblem.      To  express  the  area  of  a  triaiigle  in 
terms  of  the  three  sides  and  the  radius  of  the  circum- 
scribed circle.  ^ -^c 


In  the  figure  C^=  2r,  where  r  is  the  radius  of  the  cir- 
cumscribed circle. 

Then  ah^^r^h.  (§429) 

But  h  =  --^/s{s-a)(^s-b)(^s-c).    (§442) 

c 

4  7*      / 

Hence,  ah  =  —  Vs(s— (2)(s  — 6)(s— <?), 

c 


ahc 


or  —  =  V8(s  —  a)(8  —  6)(s  —  (?). 

4r 


But  area  of  A  ^bc  =  Vs(s  —  a)(s  —  5)(s  —  c).    (§442) 
Hence, 

areaof  A^50  =  ^- 
4r 

444.  EXERCISES. 

1.  Compute  the  areas  of  the  triangles  whose  sides  are  (1)  7,  9, 
12;   (2)  11,9,7;  (3)3,4,5. 

2.  Express  the  radius  of  the  circumscribed  circle  of  a  triangle  in 
terms  of  the  three  sides. 

3.  Find  the  area  of  an  equilateral  triangle  whose  side  is  a  by 
§  443,  and  also  without  this  theorem,  and  compare  results. 


264  PLANE  GEOMETRY. 

PROBLEMS   AND  APPLICATIONS. 

1.  Given  a  regular  dodecagon  (twelve-sided  polygon)  with  radius 
r.     Connect  alternate  vertices. 

(a)  Prove  that  the  resulting  figure  is  a  regular  hexagon. 
(h)  Find  the  apothem  of  the  hexagon. 

(c)  Find  the  area  of  each  of  the  triangles  formed  by  joining  the 
alternate  vertices  of  the  dodecagon. 

2.  Find  the  area  of  a  regular  dodecagon  of  radius  r. 

3.  Find  the  side  of  a  regular  dodecagon  of  radius  r. 

4.  Find  the  apothem  of  a  regular  dodecagon  of  radius  r, 
(a)  by  means  of  Exs.  2  and  3  and  §  343. 

(h)  by  means  of  Ex.  3  and  §  319. 

5.  Given  the  side  8  of  a  regular  dodecagon,  find  apothem.     Also 
find  the  apothem  if  the  side  is  s.    See  Exs.  9-12,  page  235. 

6.  Given  a  circle  of  rajJius  6  : 

(a)  Find  the  area  of  a  regular  hexagon  circumscribed  about  it. 
(h)  Find  the  area  of  a  regular  octagon  inscribed  in  it,  also  of  one 
circumscribed  about  it. 

7.  Using  the  formula  obtained  in  Ex.  5,  find  the  side  of  a  regu- 
lar dodecagon  whose  apothem  is  h. 

8.  Find  the  radius  of  a  circle  circumscribed  about  a  dodecagon 
whose  apothem  is  b. 

9.  Find  the  difference  between  the  radii  of  the  regular  dodecagons 
inscribed  in  and  circumscribed  about  a  circle  of  radius  10  inches. 

10.  Solve  Ex.  9  if  the  radius  of  the  circle  is  r. 

11.  What  is  the  radius  of  a  circle  if  the  difference  between  the 
areas  of  the  inscribed  and  circumscribed  regular  dodecagons  is  12 
square  inches  ? 

12.  What  is  the  radius  of  a  circle  if  the  difference  between  the 
areas  of  the  inscribed  and  circumscribed  regular  hexagons  is  8  square 
inches?    See  Ex.  11,  page  259. 

13.  A  regular  hexagon  and  a  regular  dodecagon  have  equal  sides. 
Find  these  sides  if  the  area  of  the  dodecagon  is  six  square  inches  more 
than  twice  the  area  of  the  hexagon. 

14.  Solve  Ex.  13  if  the  area  of  the  dodecagon  exceeds  twice  the 
area  of  the  hexagon  by  h  square  inches. 


REVIEW  AND   FURTHER   APPLICATIONS.  Zb5 

15.  Is  there  any  common  length  of  side  for  which  the  area  of  a 
regular  dodecagon  is  twice  that  of  a  regular  hexagon?  Three  times? 
Four  times?     Prove  your  answer. 

16.  Solve  a  problem  similar  to  that  of-  Ex.  15  if  the  side  of  tlie 
dodecagon  is  twice  that  of  the  hexagon. 

17.  The  nineteen  small  circles  in  the  accompanying  figure  are  of 
the  same  size.     The  centers  of  the  twelve  outer  circles  lie  on  a  circle 


/m| 


\oQ\ 


Yf-^f 


IS 


with  center  at  0,  as  do  the  six  circles  between  these  and  the  inner- 
most one.  The  centers  of  these  small  circles  divide  the  large  circles 
on  which  they  lie  into  equal  arcs. 

-  If  r  is  the  radius  of  the  small  circles,  then  OB  =  2h  r,  BD  =  '2h  r, 
andDG'^lir. 

(rt)    Find  A  B  and  CD  in  terms  of  i\ 

(b)    Find  DE. 

((•)  What  part  of  the  circle  OG  is  contained  within  the  nineteen 
small  circles? 

(d)  If  the  radius  of  the  large  circle  is  36  inches,  find  the  radii  of 
the  small  circles  so  that  they  shall  occupy  half  its  area. 

(e)  If  the  radius  of  the  large  circle  is  r  inches,  find  the  radii  of 
the  small  circles  so  they  shall  occupy  one  half  the  area  of  the  large 
circle. 

(/)  Under  (d)  how  far  apart  will  tlie  outer  twelve  centers  be? 
The  inner  six  ? 

(</)  Under  (e)  how  far  apart  will  the  outer  twelve  centers  be? 
The  inner  six? 

In  Exs.  (/)  and  (g)  it  is  understood  that  the  distances  are  meas- 
ured along  straight  lines. 


266 


PLANE  GEOMETRY. 


From  First  Congregational 
Church,  Chicago. 

18.  In  the  figure  ABC  is  an  equilateral  triangle.  A,B,C  are  the 
centers  of  the  arcs  BC,  CA,  and  AB.  Semicircles  are  constructed  on 
the  diameters  AB,  BC,  CA.     Let  AB  =  a.     (See  Ex.  11,  p.  93.) 

Circles  are  constructed  tangent  to  the  various  arcs  as  in  the  figure. 

(a)  Find  the  radius  r". 

Suggestion.     Find  in  order  BM,  MN,  BN,  HO",  BO". 

(b)  Find  the  radius  r'. 

Suggestion.     BN  is  known  from  the  solution  of  (a). 

BN  =  BO'  +  r'  and  BO'  =  2  KO'.     Hence  KO'  =  ^^  - ''' . 


KB  =  \y/l- BO'  =  1  V3  {BN -  r')  and  HO'  = 


But 


HO'""  =  HK''  +  KO' 


0) 


Substituting  for  HO' ,  KB  and  KO'  in  (1)  we  may  solve  for  r'. 
(c)    Find  the  radius  r. 

Suggestion.     Use  A  BLO  and  HLO  and  proceed  as  under  (i). 
{(i)  Using  the  radii  thus  found,  show  how  to  construct  the  figure, 
(e)    What  fraction  of  the  whole  area  is  contained  in  the  circles? 

19.  ABCD  is  a  parallelogram  with  fixed  base  and  altitude.  Find 
the  locus  of  the  intersection  points  of  the  bisectors  of  its  interior 
base  angles.  ^ _jp 

20.  Find  the  locus  of  a  point  P 
such  that  the  sum  of  the  squares 
of  its  distances  from  two  fixed 
points  is  constant. 

Suggestion.     By  §  440,  Zp'  +  PB^  =  2  AC^  +  2  CP\ 
Solve  for  CP. 


REVIEW  AND   FURTHER   APPLICATIONS. 


267 


21.  Find  the  locus  of  a  point  P  such  that  the  ratio  of  its  distances 
from  two  fixed  points  is  equal  to  the  constant  ratio  m  :  n. 

Suggestions.  Let  A  and 
B  be  the  fixed  points.  On  the 
line  AB  there  are  two  points 
P'  and  P"  on  the  locus,  i.e. 
AP"  :  P"B  =  AP' :  P'B  =  m:n. 

Let  P  be  any  other  point  on 
tlie  locus. 

Then  AP.BP  =  AP' :  P'B  =  A P"  :  P"B. 

Show  by  the  converses  of  §§  250,  253  that  PP'  bisects  Z  APB  and 
PP"  bisects  Z  BPK,  and  hence  P'P  and  P'<P  form  a  right  angle. 

22.  Find  the  locus  of  a 
point  P  from  which  two  circles 
subtend  the  same  angle. 

Suggestion.  C  and  C  are 
the  centers  of  the  given  circles. 
Prove  APDC^APD'C  and 
PC     r' 


hence  that 


PC 


23.   The  points  A  BCD  are  collinear.     Find  the  locus  of  a  point 
P  from  which  the  segments  AB  and  CD  subtend  the  same  angle. 

Suggestions.  The  A  ABP,  A  CP,  BDP,  and 
CDP  have  a  common  altitude,  and  hence  their 
areas  are  to, each  other  as  their  bases.  Also  A  ABP 
and  CDP  have  equal  vertex  angles,  whence  by, 
§  427  their  areas  are  to  each  other  as  the  products 
of  the  sides  forming  these  angles.     Similarly  for  ^ 


JJ    c 


D 


Hence 


AP-PB 


AB       .AP.PC 
CD  ""^  BP:i^D 


A  CP  and  BDP. 
^AC 

bd' 


CP  •  PD 

From  these  equations  show  that  AP :  PD  is  a  constant  ratio 
24.   ^^C  is  a  fixed  isosceles  triangle.     With  P 

center  C  and  radius  less  than  AC,  construct  a 
circle,  and  from  A  and  B  draw  tangents  to  it  meet- 
ing in  P.     Find  the  locus  of  P.  /\  /\  J\p 

Suggestions,  (a)  Show  that  part  of  the  locus 
is  the  straight  line  PP'.  (h)  Show  that  Z  1  =  Z  2 
and  hence  that  Z  AP"B  is  constant. 


268  PLANE  GEOMETRY. 

MAXIMA    AND  MINIMA. 

445.  Definitions.  Of  all  geometric  figures  fulfilling 
certain  conditions  it  often  happens  that  some  one  is 
greater  than  any  other,  in  which  case  it  is  called  a  maxi- 
mum. Or  it  may  happen  that  some  one  is  less  than  any 
other,  in  which  case  it  is  called  a  minimum. 

E.g.  of  all  chords  of  a  circle  the  diameter  is  the  maximum,  and  of 
all  segments  drawn  to  a  line  from  a  point  outside  it  the  perpendicular 
is  the  minimum. 

In  the  following  theorems  and  exercises  the  terms 
maximum  and  minimum  are  used  as  above  defined. 
However,  a  geometric  figure  is  often  thought  of  as  con- 
tinuously  varying  in  size.,  in  which  case  it  is  said  to  have 
a  maximum  at  any  position  where  it  may  cease  to  increase 
and  begin  to  decrease,  whether  or  not  this  is  the  greatest 
of  all  its  possible  values.  Likewise  it  is  said  to  have  a 
minimum  at  any  position  where  it  may  cease  to  decrease 
and  begin  to  increase. 

E.g.  if  in  the  figure  a  perpen- 
dicular from  a  point  in  the  curve  to 
the  straight  line  be  moved  continu- 
ously parallel  to  itself,  the  length  of 
this  perpendicular  will  have  maxima  at  A,  C,  and  E,  and  minima  at 
B,  D,  and  F. 

Certain  simple  cases  of  maxima  and  minima  problems 
have  already  been  given.  Some  of  these  will  be  recalled 
in  the  following  exercises. 

446.  EXERCISES. 

1.  If  from  a  point  within  a  circle,  not  the  center,  a  line-segment 
be  drawn  to  meet  the  circle,  show  that  this  segment  is  a  maximum 
when  it  passes  through  the  center  and  a  minimum  when,  if  produced 
in  the  opposite  direction,  it  would  pass  through  the  center. 


REVIEW  AND  FURTHER  APPLICATIONS.  269 

2.  Show  that  of  all  chords  through  a  given  point  within  a  circle, 
not  the  center,  the  diameter  is  a  maximum  and  the  chord  perpendicu- 
lar to  the  diameter  is  a  minimum. 

3.  Of  all  line-segments  which  may  be  drawn  from  a  point  outside 
a  circle  to  meet  the  circle,  that  is  a  maximum  which  meets  it  after 
passing  through  the  center,  and  that  is  a  minimum  which,  if  pro- 
duced, would  pass  through  the  center. 

4.  Show  that  if  a  square  and  a  rectangle  have  equal  perimeters, 
the  square  has  the  greater  area. 

Suggestion.     If  s  is  the  side  of  the  square  and  a  and  h  are  the 
altitude  and  base  of  the  rectangle  respectively,  then  26  +  2a  =  4sor 
_&  +  g 

HeL    .s-  ^'  +  «^+2''^  =  (V^-2ab  +  h^)+iab  ^  (h  -  oy  ^  „j_ 
4  4  4 

That  is,  s^  is  greater  than  ab  by  >^ — ^  • 

5.  Use  the  preceding  exercise  to  find  that  point  in  a  given  line- 
segment  which  divides  it  into  two  such  parts  that  their  product  is  a 
maximum. 

6.  Find  the  point  in  a  given  line-segment  such  that  the  sum  of 
the  squares  on  the  two  parts  into  which  it  divides  the  segment  is  a 
minimum. 

Suggestion.     If  a  and  b  are  the  two  parts  and  k  the  length  of  the 
segment,  then   ^^^^^  ^^^^  a^  +  2  ab  +  b^  =  k% 
or,  a2  +  ^,2  ^  ^2  _  2  ab. 

Hence,  a"^  -\-b^  is  least  when  2  ab  is  greatest.     Now  apply  Ex.  5. 

7.  In  the  preceding  exercise  show  that  a^  +  6^ 
increases  as  b  grows  smaller,  that  is,  as  the  point  of 
division  approaches  one  end.  When  is  this  sum  a 
maximum  ?  

8.  Two  points  A  and  B,  on  opposite  sides  of  a 

straight  stream  of  uniform  width,  are  to  be  con-     

nected  by  a  road  and  a  bridge  crossing  the  stream 

at  right  angles.     Find  by  construction  the  location  / 

of  the  bridge  so  as  to  make  the  total  path  from  A 

to  B  a,  minimum. 


270 


PLANE  GEOMETRY, 


maximum  area. 


447.    Theorem.      Of  all  triangles  having  equal  perim- 
eters ayid  the  same  base,  the  isosceles  triangle  has  the 

a 
c 


-ij>J^ 


A  E  B 

Given  A  ABC  isosceles  and  having  the  same  perimeter  as  A  ABD. 
To  prove  that  area  A  ABO  area  A  ABD. 

Outline  of  proof:  Draw  CE  JL  AB.  Construct  AAFB 
having  its  altitude  FE  the  same  as  that  of  A  ABD.  Pro- 
long AF  making  FG  =  AF.  Draw  GB  and  GD.  The  ob- 
ject is  to  prove  that  EF,  the  altitude  of  AABD^  is  less  than 
EC,  the  altitude  of  A  ABC. 

(1)  Show  that  A  AFB,  FBG,  and  GBD  are  all  isosceles,  for 
which  purpose  it  must  be  shown  that  GB  ±  AB  and  FD  II  AB. 

Then  AD-^  DB  =  AD-\- DG>AG. 

Or  AF-h  FB  <AD-\-  DB. 

But  AD+DB  =  AC-\-CB. 

Hence,  AF-{- FB  <AC+ CB. 

(2)  Show  that  AF<AC, 
and  hence  that  EF  <  EC. 

(3)  Use  the  last  step  to  show  that 

area  A  ABO  area  A  ABD. 
Give  all  the  steps  and  reasons  in  full. 

448.  Corollary.  Of  all  ti^iangles  having  the  same 
area  and  standing  on  the  same  base,  that  which  is 
isosceles  has  the  least  perimeter. 


REVIEW  AND  FURTHER  APPLICATIONS.  271 

449.  Theorem.  Of  all  polygons  haviiig  the  same 
perimeter  and  the  same  number  of  sides,  the  one  with 
maximum  area  is  equilateral. 

E 


Given  ABCDEF  the  maximum  of  polygons  having  a  given  perim- 
eter and  the  same  number  of  sides. 

To  prove  that  AB  =  BC  =  CD  =  DE=  EF=  FA. 

Suggestion.  Show  that  every  A,  such  as  FDE,  is  isos- 
celes by  use  of  the  preceding  theorem. 

450.  Theorem.  Of  all  jjolygons  having  the  same 
area  and  the  same  number  of  sides,  the  regular  polygon 
has  the  minimum  perimeter. 


Given  polygons  A  and  B  with  same  number  of  sides  and  equal 
area,  A  being  regular  and  B  not. 

To  prove  that  the  perimeter  of  A  is  less  than  that  of  B. 

Outline  of  Proof :  Construct  a  regular  polygon  C  having 
the  same  number  of  sides  and  same  perimeter  as  B. 

Then,  area  of  B  <  area  of  C,  or  area  of  J.  <  area  of  C. 

But  A  and  C  are  both  regular  and  have  the  same  num- 
ber of  sides.  Hence  the  perimeter  of  A  is  less  than  that 
of  C.     That  is,  perimeter  of  ^  <  perimeter  of  B, 


272  PLANE  GEOMETRY, 

451.  Theorem.  Tlie  polygon  with  maximum  area 
which  can  he  formed  hy  a  series  of  line-segments 
of  given  lengths^  starting  and  ending  on  a  given  line, 
is  that  one  ivhose  vertices  all  lie  on  a  semicircle  con- 
structed on  the  intercejoted  part  of  the  giveji  line  as  a 
diameter. 


B- 


F 


Given  the  line-segments  AB,  BC,  CD,  DE  and  EF  meeting  the 
line  RS  at  A  and  F  so  as  to  form  the  polygon  ABCDEF  with  maxi- 
mum area. 

To  prove  that  jB,  C,  d,  and  E  lie  on  the  semicircle  whose 
diameter  is  AF. 

Proof:  Suppose  any  vertex  as  B  does  not  lie  on  this 
semicircle.  Join  D  to  F  and  to  A.  Then  Z  ADF  is  not  a 
right  angle,  else  it  would  be  inscribable  in  a  semicircle' 
(§  213). 

If  now  the  extremities  A  and  F  be  moved  in  or  out  on 
the  line  liS^  keeping  AD  and  DF  unchanged  in  length,  till 
Z  ADF  becomes  a  right  angle,  then  A  ADF  will  be  increased 
in  area  (Ex.  3,  §  456),  while  the  rest  of  the  polygon  is 
unchanged  in  area.  This  would  increase  the  total  area  of 
the  polygon  ABCDEF^  which  is  contrary  to  the  hypothesis 
that  this  is  the  polygon  with  maximum  area.  Hence  the 
vertex  D  must  lie  on  the  semicircle. 

In  the  same  manner  it  can  be  proved  that  each  vertex 
lies  on  the  semicircle. 


REVIEW  AND  FURTHER  APPLICATIONS. 


273 


452.  Theorem.  Of  all  polygons  with  the  same  mim- 
&er  of  sides  equal  m  pairs  and  taken  in  the  same  order, 
the  one  ivith  maximum  area  is  that  one  vjhich  can  he 
inscribed  in  a  circle. 


K' 


-ELkil 


Given  a  polygon  ABCDE  inscribed  in  a  circle  and  A'BC'D'E' 
not  inscribable  but  with  sides  respectively  equal  to  those  of 
ABCDE. 

To  prove  that       ABCDE>a'b'c'd'e'. 

Proof:  In  the  first  figure  draw  the  diameter  DK  and 
draw  AK  and  BK. 

On  a'b'  construct  A  a'b'k  ^  Aabk  and  draw  d'k'. 

The  circlte  whose  diameter  is  b'k'  cannot  pass  through 
all  the  points  A',  b',  c\  and  E\  else  the  polygon  would  be 
inscribed  contrary  to  hypothesis. 

If  either  A'  or  E'  is  not  on  this  circle,  then 

AKBE>a'k'd'e'.  (Why?)     (1) 

Likewise  if  either  b'  or  &  is  not  on  the  circle,  then 

KBCD  >  k'b'c'b'.  (Why  ?)     (2) 

In  any  case,  adding  (1)  and  (2), 

AKBCDE  >  A'e'b'c'd'e'. 
By  construction,    A  AKB  ^  A  A' k' b' . 
From  (3)  and  (4),  ABODE  >  a'b'c'd'e'. 


(3) 
(4) 


274 


PLANE  GEOMETRY. 


453.  Theokem.  Of  all  regular  polygons  having 
equal  perimeters,  that  which  has  the  greatest  number 
of  sides  is  the  maximum. 

F 


Given  T  a  regular  triangle  and  S  a  square  having  the  same 
perimeter. 

To  prove  that  s  >  T. 

Proof :  Take  any  point  E  in  AB.  Draw  CE  and  on  CE 
construct  A  CEF  ^  A  A  EC. 

Then  polygon  BCFE  has  a  perimeter  equal  to  that  of  T 
and  the  same  area,  and  has  the  same  number  of  sides  as  -s. 
Hence,  s  >  BCFE.  (§  449) 

That  is,  s  >  T. 

In  like  manner  it  can  be  shown  that  a  regular  polygon 
of  five  sides  is  greater  than  a  square  of  equal  perimeter, 
and  so  on  for  any  number  of  sides. 

454,  EXERCISE 

Of   the   three   medians  of    a   scalene     C 
triangle  that  one  is  the  shortest  which  is 
drawn  to  the  longest  side.  F 

Suggestion.     If  AC<BC,  to  show 
th&t  BF>AE.  A  D  B 

Jn  A  A  DC  and  BDC,  AD  =  DB  and  DC  is  common. 

But^C<^C.     Hence  Z1<Z2  (Why?). 

Now  use  A  ADO  and  BDO  to  show  that  BO  >  .4  0  and  hence  that 
BF>AE. 


E 


REVIEW  AND  FURTHER   APPLICATIONS. 


275 


455.  Theorem.  Of  all  regular  polygons  having  a 
given  area,  that  one  which  has  the  greatest  number  of 
sides  has  the  least  perimeter. 


Given  regular  polygons  P  and  Q  such  that  P  =  Q  while  Q  has  the 
greater  number  of  sides. 

To  prove  that  perimeter  of  Q  <  perimeter  of  P. 

Proof :    Construct  a  regular  polygon  R  having  the  same 
number  of  sides  as  P  and  the  same  perimeter  as  Q. 
Then  r  <q.  (§453) 

But  P  =  Q.  (By  hypothesis) 

Hence  R  <  P. 

Therefore,  perimeter  oi  R  <  perimeter  of  P.  (§  323) 

But  R  was  constructed  with  a  perimeter  equal  to  that  of  Q. 
Hence  perimeter  of  Q  <  perimeter  of  P. 


456. 


EXERCISES. 


1.  Of  all  inscribed  polygons  of  a  given  number  of  sides  in  a  given 
circle,  that  one  which  is  regular  has  the  maximum  area  and  the  maxi- 
mum perimeter. 

2.  Of  all  circumscribed  polygons  of  a  given  number  of  sides  about 
a  given  circle,  that  one  which  is  regular  has  the  least  perimeter  and  the 
least  area. 


3.  Of  all  triangles  ,  having  two  given 
sides,  that  in  which  these  sides  include  a 
right  angle  has  the  maximum  area. 

Suggestion.  Show  that  the  altitude  is 
greatest  when  the  angle  is  a  right  angle. 


Ci 


INDEX. 


(References  are  to  sections  unless  otherwise  stated.) 


Abbreviations 80 

Acute  angle 19 

triangle 23 

Addition  of  angles 39 

of  line-segments 10 

Adjacent  angles 39 

parts  of  a  polygon 134 

parts  of  a  triangle 21 

Alternate  exterior  angles     .     .  88 

interior  angles 88 

Alternation 24G 

Altitude  of  a  parallelogram  .     .  137 

of  a  trapezoid 137 

of  a  triangle 24 

Analysis  of  proof 175 

of  problems     .......  177 

Angle,    acute,     obtuse,     right, 

reflex,  oblique 19 

at  the  center  of  a  circle   .     .     .  18o 

exterior  of  a  triangle  ....  81 

inscribed  in  a  circle     ....  185 

inscribed  in  a  segment,  an  arc  .  220 
measurement  of  .     .     .     .     .17, 202 

reentrant 134 

straight,  zero,  perigou     .     .     10,  17 

vertex  of,  side  of 14 

Angles,  adjacent 39 

addition  of,  sum  of      ....  39 

complementary 70 

multiplication  and  division  of  .  39 

subtraction  of 39 

supplementary 70 

vertical 70 

Antecedents 242 

Apothem  of  a  regular  poly- 
gon      'M2 

Approximate  area 306 

measurement      ....      235,  305 


Approximate  ratio      ....    240 

Arc  of  a  circle 12,  184 

intercepted  by  an  angle,  .  .  .  185 
subtended  by  a  choril  ....     184 

Area,  approximate 306 

of  a  circle 'Ml,  416 

of  a  polygon :W9 

of  a  rectangle  .  .  .  300-307,  414 
of  a  sector  of  a  circle  ....  364 
of  a  segment  of  a  circle   .     .     .    365 

of  a  surface 302 

Aristotle page  31 

Axial  symmetry 164 

Axioms  .  60,61,82,96,119,392, 

407,409,411,412,416 

on  numbers 241 

Axis  in  a  graph ^^68 

of  symmetry 164 

Base  of  a  parallelogram     .  " .     .  135 

of  a  trapezoid 136 

of  a  triangle 24 

Bhaskara 318 

Bisector  of  an  angle     ....  19 

of  a  line-segment 50 

Broken  line 11 

Center  of  a  circle 12 

of  a  pencil 431 

of  a  regular  polygon    ....    342 

of  similitude 434 

of  symmetry 169 

Central  angle 185 

symmetry 169 

Chord  of  a  circle 181 

Circle,  arc  of 184 

area  of 361,  416 

circumference^of     .     .'    .      352,  416 


277 


278 


INDEX. 


(References  are  to  sections  unless  otherwise  stated.) 


Circle,  circumscribed    ....  225 

definition  of 12 

diameter  of 181 

escribed 229,  Ex.  4 

inscribed 228 

inside  of,  outside  of 180 

length  of,  perimeter  of    .    .  352,  416 

secant  of 182 

segment  of 220 

tangent  to 183 

Circles,  equal 187 

intersecting 186 

tangent 186 

Coincidence 19,  27 

Commensurable  segments   .  236 

ratios '238 

Compasses 33 

Complementary  angles     .     .  70 

Composition 246 

and  division 246 

Concave  polygon 160 

Conclusion 78 

Concurrent  lines 228 

Congruence 27 

Consequents 242 

Constants 380 

Construction  of  figures   ...  44 

of  regular  polygons     .     .     .  163,  333 

Continued  proportion   .     .     .  267 

Converse  theorems   ....  98 

Convex.polygon 160 

Corollary 104 

Corresponding  angles     ...  88 

parts  of  polygons 255 

parts  of  triangles 29 

Curved  line 11 

Decagon 424 

Definition,  technical     ....  .389 

Degree  of  an  angle    ....  17 
Demonstration  .     .      59,  79,  80,  391 

type  of   .     .    ■.     . 83 

Dependence  of  variables  .     .  380 
Diagonal  of  a  polygon  .     .  134,  Wd 

Diameter  of  a  circle     ....  181 

of  a  parallelogram 137 

of  a  trapezoid 137 

Direct  proof 175,  399 

Direct  tangent  to  tw^o  circles  403 


Discussion     of     a    construc- 
tion      122 

Distance  between  two  points    .  61 

from  a  point  to  a  line       .     .     .  114 

Dividers 33 

Division  of  line-segments      .     .  10 

external,  internal 425 

harmonic 425 

in  extreme  and  mean  ratio  .     .  421 

in  proportion 246 

of  angles 39 

of  the  circle 333 

Dodecagon      ....  pages  82, 235 

End-point  of  a  segment    .     .     .  8,  13 

of  a  ray 9 

Equal,  equivalent 27 

circles 187 

Equiangular  figures    ...  22,  160 

mutually 255 

Equilateral  figures  .  .  ,22,  160 
Euclid      .     .     .      pages  3, 31,40,  141 

Eudoxus page  140 

Exterior  angle  of  a  triangle  .     .      81 

angles  of  parallel  lines  ...  88 
External  division      .     .     .      252,  425 

segment  of  a  secant  ....  274 
Extreme  and  mean  ratio  .  ,  421 
Extremes 242 

Figures,  geometric  .....  12 

congruence  of 27 

constructions  of 44 

symmetry  of 164 

Fourth  proportional       .    ,     .  291 

Geometric  proposition  ...      59 

Geometric  solid 3 

Geometry 1 

Gothic  arch,  pages  109,  151,  197,  204 
Graphic  representation. 

Chapter  VI 

Harmonic  divisions  ....  425 
Hexagon,  regular,  pages  (55,  67, 

76,  78,  82,  108,  174,  177,  etc. 
Historical  notes,  pages  3,  31, 

38,  40,  42,  128,  140,  162,  167,  193 

Hypotenuse 23 

Hypothesis 78 


INDEX. 


279 


(References  are  to  sections  unless  otherwise  stated.) 


Incommensurables,    236,  240, 

305,  368 
Indirect  proof  ...  91,  176,  394 
Inscribed  angle  in  an  arc      .    .    220 

angle  in  a  circle 185 

angle  in  a  segment       ....    220 

circle  in  a  triangle 228 

triangle  in  a  circle       ....    225 

Intercepted  arc 185 

Interior  angles 88 

Internal  division    .    .     .      252,425 

Inversion 246 

Isosceles  trapezoid 136 

triangle 22 

Land  surveying      .     .     .     283, 284 

Legs  of  a  right  triangle     .     .      23 

Length    3 

of  a  circle 352,  416 

of  a  segment 232,  407 

Limit  of  apothem 383 

of  area 384 

of  perimeter 384 

of  rotating  secant   .....    386 

Line J,  3,  4,  5,  7,  11 

half-line 9 

of  centers 403 

-segment 8 

Lines,  concurrent 228 

oblique,  obtuse,  perpendicular  .      19 
parallel       89 

Lobachevsky      ....      page  38 

Loci,  determination  of, 

pages  50-52,  226-227 
problems,  pages  52,  64,  83,  92, 
96,  97,  102,  103,  111,  120,  148, 
149,  158,  175,  179,  226,  228,  233,  etc. 

Major  and  minor  arcs  ...  184 
Maxima  and  minima  .  .  .  443 
Mean  proportional     ....    293 

Means 242 

Measurement,  approximate,  235, 306 

of  angles 17,202 

of  line-segments 232-237 

of  the  circle 349 

exact 232,301,304 

Median  of  a  triangle  ....  157 
Methods  of  attack     .     .     .  174-177 


Minute  of  angle 17 

Multiplication  of  angles  ...      39 
of  line-segments 10 

Number 241,  407 

Numerical  measure  .  232,  302,  407 

Oblique  lines,  angles    ....      19 

Obtuse  triangles 23 

Octagons    .     .     .     pages  78, 79,  etc. 
regular  .     .  pages  146,  147,  187, 

198,  202,  235,  etc. 
Opposite  parts  of  figures  ,  21,  134 

Origin  of  ray 9 

Outline  of  proof 52 

Parallel  lines 89 

axiom 96 

rulers 288 

Parallelogram 135 

Pencil  of  lines 431 

Pentagon 424 

Perigon 17 

Perimeter  of  a  circle    ....    352 

of  a  polygon 160,  267 

Perpendicular 19 

bisector  of  a  segment  ....  50 
n  (pi),  computation  of .  .  .  .  351 
Plane,  plane  surface      .     .     .     .    1, 2 

Plato page  140 

Point 1,  3 

Polygons 160. 

regular 332 

similar 256 

Practical  measurement      235,  306 
Preliminary  theorems, 

(52-69,  72-76,  189-198,  307 
Problems  of  construction      .      44 

Projection 436 

Proof 59,  391,  392 

by  exclusion 86 

direct,  indirect    .     .  91,  175, 176,  394 

Proportion 242 

Proportional  division     .    .     .    242 
Proposition,  geometric     ...      59 

Protractor 18,  33 

Pythagoras     .    .  pages  40, 130, 

140,  162,  167 


280 


INDEX, 


(Keferences  are  to  sections  unless  otherwise  stated.) 


Quadrant 208 

Quadrilaterals 134-137 

Radius  of  circle 12 

of  regular  polygon 342 

Ratio 238,242,408 

approximate 240 

commensurable 239 

of  similitude 434 

Ray 9 

Rectangle 135 

Reflex  angle 19 

Regular  polygon 160 

Rhomboid,  rhombus     ....  135 

Right  angle 19 

triangle 23 

Scalene  triangle 22 

Secant 182 

limiting  position  of      ....     386 
Second  of  angle 17 

of  arc     .     .^ 202 

Sector  of  a  circle 203 

area  of ^564 

Segment  of  a  circle 220 

of  a  line 8 

Semicircle 184 

Sides  of  an  angle    ....   14,  106 

Similar  figures 27,  256 

Sine  of  an  angle 279 

Solution  of  problems      .     .  177-178 

Square 135 

Subtended  angle 286 

arc ■.     .     184 

Subtraction  of  angles  ....      39 

of  line  segments 10 

Sum  of  angles 39 

Summaries,  pages  75,  10(5,  145, 

169.  194 


Superposition     ....      page  14 
Supplementary  angles  ...      70 

Surface 2 

Symbols  of  abbreviation  .     .      80 

Symmetry 164-173 

problems  involving,     pages  81, 
82,  93,  94,  101,  109,  150,  205, 

229,  233,  etc. 

Tangent  to  a  circle 183 

circles 186 

to  two  circles 403,  404 

Technical  words 389 

/Tests  for  congruence     .    32,  35,  40 

xThales pages  3,  17 

Theorem 59 

Third  proportional     ....    293 

Transversal 88 

Transverse  tangent    ....    403 

Trapezoid 136 

Triangle    .    21-24,  41,  157,  225, 

228,' etc. 

Undefined  words 389 

Unproved  propositions     .  60,  391 

Variables Chapter  VI 

dependence  of 380 

limit  of 384 

Variation,  continuous  ....  366 

Vertex  angle  of  a  triangle  .     .  24 

Vertical  angles 70 

Vertices  of  a  polyuon  ....  160 

of  a  quadrilateral 134 

of  a  triangle 21 

Whole  secant 274 

Zero  angle 16 


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